Algebraic Expression, Identities And Factorisation

⇒ Monomial has single term while binomial has two terms.

⇒ Let us suppose a monomial = c and a binomial = (a – b)

⇒ product of ‘c’ and (a – b) are as follows:

⇒ c× (a – b)

= c× a – c× b

= (ca – cb) (This is a binomial)

⇒ Thus we can conclude that product of a monomial and a binomial is binomial.

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

⇒ according to definition Polynomial is defined as an expression consisting of variables and coefficients that involves non – negative exponents of variable.

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does match to our solution.

Option (d) does not match to our solution.

So, option (c) is correct answer.

⇒ (a – b)² =(a – b)(a – b)

=(a× a – a× b – a× b + b× b)

=(a² – 2ab + b²)

Option (a) does match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (a) is correct answer.

The addition is as follows:

⇒ -7pq + 2pq

=pq(-7 +2)

=pq(-5)

= (-5pq)

⇒ Result is as follows: (-5pq)

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does match to our solution.

So, option (d) is correct answer.

Subtraction is as follows:

⇒ x²y² – (-3x²y²)

= x²y² + 3x²y²

= 4x²y²

The result is as follows: 4x²y²

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does match to our solution.

So, option (d) is correct answer.

⇒ Like term are the terms which have same variable and powers. The coefficients need not to be matched.

⇒ Here 4m³n² and – 6m³n² are like term. Because they have same variables and powers.

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option B. is correct answer.

⇒ A binomial is a polynomial which have two terms.

⇒ in option (a)

⇒ given polynomial = (7× a + a)

=7a +a

=8a

⇒ 8a is monomial.

⇒ in option (b)

⇒ Given polynomial = 6a² + 7b + 2c

⇒ There are three terms in this polynomial. So, this is a trinomial.

⇒ in option (c)

⇒ Given polynomial = 4a× 3b× 2c

=24abc

⇒ There are one term in this polynomial.

∴ This polynomial is a monomial.

⇒ in option (d)

⇒ Given polynomial = 6(a² + b)

= 6a² +6b

⇒ there are two terms in this polynomial.

∴ this polynomial is binomial.

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does match to our solution.

So, option (d) is correct answer.

Addition is as follows:

⇒ (a – b + ab)+(b + c – bc)+(c – a – ac)

= (a – b + ab + b + c – bc + c – a – ac)

=(2c + ab – bc – ac)

=(2c + ab – ac – bc)

Option (a) does match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (a) is correct answer.

The multiplication is as follows:

⇒ (4p)× (-7q³)× (-7pq)

= (-28pq³)× (-7pq)

= (196p² q⁴)

Option (a) does match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (a) is correct answer.

⇒ Area of a rectangle = length× width

⇒ Area of a rectangle = 4ab× 6b²

⇒ Area of a rectangle = 24ab³

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

⇒ Volume of rectangular box (cuboid) = length× breadth× height

⇒ Volume of rectangular box (cuboid) = 2ab× 3ac× 2ac

⇒ Volume of rectangular box (cuboid) = 12a³bc²

Option (a) does match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (a) is correct answer.

The multiplication is follows:

⇒ (6a² – 7b + 5ab) × 2ab

=(6a²× 2ab – 7b× 2ab + 5ab× 2ab)

=(12a³b – 14ab² + 10 a²b²)

⇒ Result is as follows: (12a³b – 14ab² + 10 a²b²)

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

⇒ To find square of (3x – 4y) using formula –

⇒ (a – b)²=a² + b² – 2ab

⇒ here, a = 3x and b= 4y

⇒ (3x – 4y)²= (3x)² + (4y)² – 2× (3x)×(4y)

⇒ (3x – 4y)²= 9x² + 16y² – 24xy

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does not match to our solution.

Option (d) does match to our solution.

So, option (d) is correct answer.

⇒ Like term are the terms which have same variable and powers. The coefficients need not to be matched.

⇒ Here -5xyz² and 7xyz² have same variables and same powers.

∴ - 5xyz² and 7xyz² are like terms.

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

⇒ As we know that coefficients are the numbers which multiplies the variables.

⇒ In the given term , we can see that ‘y’ is a variable and is a constant which multiplies variable ‘y’.

⇒ In mathematical form –

⇒

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does match to our solution.

Option (d) does not match to our solution.

So, option (c) is correct answer.

⇒ Solving option (a)

⇒ (a – b)² = (a – b)× (a – b)

⇒ =(a× a – a× b – b× a + b× b)

=(a² – 2ab + b²)

⇒ Solving option (b)

⇒ (a – b)(a – b) = (a× a – a× b – b× a + b× b)

⇒ =(a² – 2ab + b²)

⇒ Solving equation (c)

⇒ (a + b)(a – b) = (a× a – a× b + b× a – b× b)

⇒ = (a² – b²)

⇒ Solving option (d)

⇒ (a + b)(a + b) = (a× a + a× b + b× a + b× b)

⇒ = (a² +2ab + b²)

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

⇒ Factorise the given terms

⇒ 17 abc = 17× a× b× c

⇒ 34 ab = 17× 2× a× b× b

⇒ 51ab = 17× 3× a× a× b

⇒ Common factor = 17× a× b

⇒ Common factor = 17ab

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

⇒ Applying formula

⇒ (a – b)² = a² + b² – 2ab

⇒ square of (9x – 7xy) means (9x – 7xy)²

⇒ Here, a = 9x and b = 7xy

⇒ (9x – 7xy)² = (9x) + (7xy) – 2×(9x)×(7xy)

⇒ (9x – 7xy)² = 81x² + 49x²y² – 126x²y

Option (a) does not match to our solution.

Option (b) does not match to our solution.

Option (c) does match to our solution.

Option (d) does not match to our solution.

So, option (c) is correct answer.

lets write factorise form of 23xy – 46x + 54y – 108

⇒ 23xy – 46x + 54y – 108 = 23x(y – 2) + 54(y – 2)
Taking (y-2) as common from the term, we get,

= (y – 2)(23x + 54)

= (23x + 54)(y – 2)

Hence, option (a) is correct

⇒ Factorise form will be as follows:

⇒ (r² -10r + 21) = r – (7+3)r + 21

⇒ (r² -10r + 21) = r – 7r – 3r + 21

⇒ (r² -10r + 21) = r(r – 7) – 3(r – 7)

⇒ (r² -10r + 21) = (r – 7)(r – 3)

Option (a) does not match to our solution.

Option (b) does match to our solution.

Option (c) does not match to our solution.

Option (d) does not match to our solution.

So, option (b) is correct answer.

We have,

(By splitting the middle term, so that the product of their numerical coefficients is equal constant term)

=p(p-19)+2(p-19)

=(p-19)(p+2)

(Because )

Required value

We have,

(Because )

We have monomials 3ab and 2cd

Now,

We see that, there is no common factor between them except 1

(Neither numerical nor letter)

A factor is said to be irreducible, if it cannot be factorised further

We have, 24x²y² = 2×2×2×3×x×x×y×y

Hence, an irreducible factor of is

We can write as and this cannot be

Factorised further.

Hence, number of factors of is 2

(Taking 3 as common)

(Because )

Hence, , are factors of

We have,

=3x

(d)

Given: (3x³ +9x² + 27x) ÷ 3x

= x² + 3x + 9

C

Given:

(a + b)² + (a – b)²

⇒ a² + b² + 2ab + a² + b² – 2ab

⇒ 2a² + 2b²

(a + b)² - (a – b)²

⇒ a² + b² + 2ab - a² - b² + 2ab

⇒ 4ab

Positive

If both like terms are either positive or negative, then the resultant term will always be positive.

Negative

As the product of a positive term and a negative term is always negative.

b,c

We have,

(Using left distributive law => The law relating the operations of multiplication and addition, stated symbolically, a(b+c)=ab+ac; that is, the monomial facto a is distributed, or separately applied, to each term of the binomial factor b+c, resulting in the product ab+ac. From this law it is easy to show that the result of first adding several numbers and then multiplying each separately by the number and then adding the products)

(a-b)

We know that,

(Because )

(a-b)

We have,

(Because )

Alternative method

Let

Let

(Because )

We have,

(Because )

ab

polynomial

An expression containing one or more terms with non-zero coefficient, with variables having non-negative exponents, is called a polynomial. When we multiply two polynomials, the coefficients of the terms get multiplied and the power of the variables get added. This gives us the resulting answer as a polynomial, too.

x

In both of the above terms, the common factor is .

2mn(9 + 5p)

The irreducible factors of 18mn and 10mnp are

18mn = 2×3×3×m×n

10mnp = 2×5×m×n×p

Here, the common factors of the two terms are 2mn. So the factorization of 18mn + 10mnp will be

18mn + 10mnp = (2×3×3×m×n) + (2×5×m×n×p)

= (2mn×9) + (2mn×5p)

= 2mn(9 + 5p)

(2y - 3)²

Here, the polynomial has three terms where the first and the last term are perfect squares with a negative sign of the middle term. So it is of the form where a = 2y and b = 3 such that

Now, we know that

Comparing with this, we get,

2x²z

While dividing two monomials, the coefficient of the dividend is divided by the coefficient of the divisor whereas the variables of the dividend are divided by the variables of the divisor.

Here, dividend =

divisor =

So,

Hence, the answer is

24xyz

given:- length of box = 2x

breadth of box = 3y

height of box = 4z

Volume of rectangular box = length × breadth × height

= (2x) × (3y) × (4z)

= 24xyz

(67 + 37), 3120

67² – 37² is of the form a² – b². Comparing the two we get, a = 67 and b = 37.

We have a² - b² = (a + b)(a-b).

So, 67² – 37² = (67 + 37)(67-37)

= (104)(30)

= 104×30 = 3120

(103 + 102), 205

103² – 102² is of the form a² – b². Comparing the two we get, a = 103 and b = 102.

We have a² - b² = (a + b)(a-b).

So, 103² – 102² = (103 + 102)(103-102)

= (205)(1)

= 205×1 = 205

12x²y²

Let the length of rectangular plot = l = 4x²

The breadth of rectangular plot = b = 3y²

Area of rectangular plot = l×b

= 4x² × 3y²

= 12x²y²

8x³

Given:- l = b = h = 2x

Volume of rectangular plot = l×b×h

= 2x × 2x × 2x

= 8x³

-37

The numerical factor of a term is called its coefficient. Here, the numerical factor is -37.

two

The expression can be written as

A² + bc×d = a² + bcd

The multiplication sign does not mean that there are two separate terms. Two terms can be separated only by the operations of addition or subtraction.

16(a² + b²)

Given:- side of first square = 4a

⇒ Area of first square = (side)²

= (4a)²

= 16a²

side of second square = 4b

⇒ Area of second square = (side)²

= (4b)²

= 16b²

Sum of areas of the two squares = 16a² + 16b²

= 16(a² + b²)

distributive

3y

Area of a square = (side)²

⇒ 9y² = (side)²

Taking square root on both sides, we get,

side =

To find the square root of a monomial, we find the square roots of the coefficient and variables separately.

Now, 9y² = (3y)²

Thus, side of square = 3y

We have,

2(x + 2y)

2x + 4y = (2×x) + (2×2×y)

= (2×x) + (2×2y)

= 2(x + 2y)

Lets recall the formula of (a + b)².

(a + b)² = a² + 2×a×b + b²

⇒ (a + b)² = a² + 2ab + b²

But right-hand side of the equation does not match left-hand side of the equation.

That is,

a² + 2ab + b² ≠ a² + b²

⇒ (a + b)² ≠ a² + b²

Hence, this statement is false.

Expansion of (a – b)² is,

(a – b)² = a² – 2×a×b + b²

⇒ (a – b)² = a² – 2ab + b²

And a² – 2ab + b² ≠ a² – b²

⇒ (a – b)² ≠ a² – b²

Hence, this statement is false.

Let us expand (a + b)(a – b).

Multiply each term with the other term which is in multiplication with it.

(a + b)(a – b) = a² – ab + ab – b²

⇒ (a + b)(a – b) = a² +(ab – ab) – b²

⇒ (a + b)(a – b) = a² + 0 – b²

⇒ (a + b)(a – b) = a² – b²

Hence, this statement is true.

Let us understand this by an example.

Let two negative terms be -a and -b.

Multiplying these two negative terms.

-a × -b = (-1) × (-1) × ab

⇒ -a × -b = +1 × ab [∵, -1 × -1 = 1]

⇒ -a × -b = +ab

The product of -a and -b came out to be ab (positive term).

Whenever two negative terms are multiplied, the product is always positive as the two negatives forms a positive.

Hence, this statement is false.

Let us understand this by an example.

Let a negative and a positive term be -a and b respectively.

Then, multiply these two terms.

-a × b = (-1) × (+1) × ab

⇒ -a × b = -1 × ab [∵, -1 × 1 = -1]

⇒ -a × b = -ab

Similarly, take any pair of such numbers, we will always get a negative term as a negative number multiplied by a positive number always gives negative number.

Hence, this statement is true.

Coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression.

The algebraic expression in -6x²y² is x²y².

And the constant term before x²y² is -6.

Thus, coefficient of the term -6x²y² is -6.

Hence, the statement is true.

An algebraic expression which consists of two non-zero terms is called a binomial.

Observe in the given question, there are three non-zero terms, namely

p²q, r²r and r²q.

Thus, this is a polynomial having three terms, not a binomial.

Hence, the statement is false.

Lets factorize a² – 2ab + b² by splitting middle term.

a² – 2ab + b² = a² – ab – ab + b² [∵, -ab – ab = - 2ab & (-ab)(-ab) = a²b²]

⇒ a² – 2ab + b² = a(a – b) – b(a – b)

⇒ a² – 2ab + b² = (a – b)(a – b) = (a – b)²

So, the factors of a² – 2ab + b² are (a – b) and (a – b).

Hence, the statement is false.

Factorizing 2π (h + r), we get

2π (h + r) = 2π × (h + r)

⇒ There are three factors of 2π (h + r) namely, 2, π and (h + r).

But h is not any factor of 2π (h + r).

Hence, the statement is false.

Factorizing n²/2 + n/2, by simply taking out common variables and constants.

n²/2 + n/2 = n²×1/2 + n×1/2

⇒ n²/2 + n/2 = 1/2 (n×n + n) = 1/2 n (n + 1)

⇒ 1/2, n and (n + 1) are factors of n²/2 + n/2.

Hence, the statement is true.

An equation is a statement of equality between two quantities or algebraic expressions. Most algebraic equations are TRUE when certain values are substituted for the variable (such as x) and are FALSE for all other values. The values that make equations TRUE are called "solutions". So, an equation is not necessarily true for all values of its variables.

Hence, the statement is false.

We need to factorize x² + (a + b)x + ab, by splitting the middle term:

x² + (a + b)x + ab = x² + ax + bx + ab [∵, ax + bx = (a + b)x & ax × bx = abx²)

⇒ x² + (a + b)x + ab = x(x + a) + b(x + a)

⇒ x² + (a + b)x + ab = (x + a)(x + b)

And (x + a)(x + b) ≠ (a + b)(x + ab)

⇒ x² + (a + b)x + ab ≠ (a + b)(x + ab)

Hence, the statement is false.

Common factors of 11pq², 121p²q³ and 1331p²q can be found out as:

Common factors (11pq², 121p²q³, 1331p²q) = Common factors (11×pq, 11×11p²q³, 11×121p²q)

⇒ Common factors (11pq², 121p²q³, 1331p²q) = 11

[∵, 11 divides all the three term 11pq², 121p²q³ and 1331p²q; p divides all the three terms 11pq², 121p²q³ and 1331p²q; and q divides all the three terms 11pq², 121p²q³ and 1331p²q]

Hence, the statement is false.

Let us consider that common factor of 12a²b² + 4ab² – 32 = 4

Check: 12a²b² + 4ab² – 32 = 4 (3a²b² + ab² – 8)

Notice, there are no further common factor in the equation.

Thus, common factor of 12a²b² + 4ab² – 32 is 4.

Hence, the statement is true.

Let us factorize -3a² + 3ab + 3ac.

-3a² + 3ab + 3ac = 3(-a×a + a×b + a×c)

⇒ -3a² + 3ab + 3ac = 3a (-a + b + c)

But, 3a (-a + b + c) ≠ 3a (-a – b – c)

Hence, the statement is false.

Factorizing the equation p² + 30p + 216, by splitting the middle term of the equation:

p² + 30p + 216 = p² + 18p + 12p + 216 [∵, (18p + 12p) = 30p & (18p × 12p) = 216p²)

⇒ p² + 30p + 216 = p (p + 18) + 12 (p + 18)

⇒ p² + 30p + 216 = (p + 12) (p + 18)

But, (p + 12)(p + 18) ≠ (p + 18)(p – 12)

Hence, the statement is false.

Let two consecutive numbers be x and (x + 1).

Then square of these numbers are x² and (x + 1)².

Difference of squares of these consecutive numbers = (x + 1)² – x²

= x² + 1 + 2x – x² [∵, (a + b)² = a² + b² + 2ab)

= 2x + 1

= x + x + 1

= (x) + (x + 1)

= sum of the two consecutive numbers x and x+1

Thus, difference of squares of two consecutive numbers = sum of the same consecutive numbers.

Hence, the statement is true.

An algebraic expression which consists of one non-zero terms is called a monomial.

Observe in the given question, there are three non-zero terms, namely

abc, bca and cab.

Thus, this is a polynomial having three terms, not a monomial.

Hence, the statement is false.

Divide p/3 by 3/p, we get

p/3 ÷ 3/p = p/3 × p/3 = p²/9

And p²/9 ≠ 9.

⇒ the quotient is p²/9, not 9.

Hence, the statement is false.

Solving for p,

(51)² – (49)² = 100p

⇒ (50 + 1)² – (50 – 1)² = 100p [∵, 51 = 50 + 1 & 49 = 50 – 1]

⇒ (2500 + 1 + 100) – (2500 + 1 – 100) = 100p [∵, (a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab)

⇒ 2500 + 1 + 100 – 2500 – 1 + 100 = 100p

⇒ 200 = 100p

⇒ 100p = 200

⇒ p = 200/100 = 2

The value of p is 2.

Hence, the statement is true.

Solving (9x – 51) ÷ 9,

(9x – 51) ÷ 9 = (9x – 51)/9

9 divides the term 9x but does not divide 51.

So, the answer comes out to be (9x – 51)/9.

(9x – 51)/9 ≠ x – 51

Hence, the statement is false.

Solving it, we get

(a + 1)(a – 1)(a² + 1) = {(a + 1)(a – 1)} × (a² + 1)

⇒ (a + 1)(a – 1)(a² + 1) = (a² – 1)(a² + 1)

[∵, (x + y)(x – y) = x² – y²; Put x = a and y = 1 ⇒ (a + 1)(a – 1) = a² – 1]

⇒ (a + 1)(a – 1)(a² + 1) = a⁴ – 1

[∵, (x² + y²)(x² – y²) = x⁴ – y⁴; Put x = a and y = 1 ⇒ (a² + 1)(a² – 1) = a⁴ – 1]

Thus, the value of (a + 1)(a – 1)(a² + 1) is a⁴ – 1.

Hence, the statement is true.

The addition is as follows:

The result is as follows 10a²bc – abc²

The addition is as follows:

The result is as follows 10ax-2by + 2cz

The addition is as follows:

The result is as follows 4xy²z²-6x²y²z-3x²yz²

: The addition is as follows:

The result is as follows 7x² + 2xy + 11y² + 4

The addition is as follows:

The result is as follows -p⁴-4p³-2p²-6p-5

To find: 3a(a - b - c) + 2b(a - b + c)
The addition is as follows:

Here,
3a(a - b - c) = 3a² - 3ab - 3ac

2b(a - b + c) = 2ab - 2b² + 2bc

Hence the two polynomials to be added are 3a² - 3ab - 3ac and 2ab - 2b² + 2bc

3a(a - b - c) + 2b(a - b + c) = 3a² - 3ab - 3ac + 2ab - 2b² + 2bc = 3a² - ab - 3ac + 2bc - 2b²

The result is as follows: 3a² - ab - 3ac + 2bc - 2b²

Here, 3a(2b + 5c) = 6ab + 15ac

3c(2a + 2b) = 6ac + 6bc

Hence the two polynomials to be added are 6ab + 15ac and

6ac + 6bc

The result is as follows 6ab + 21ac + 6bc

The subtraction is as follows:

The result is as follows:- 12a²b²c²

The subtraction is as follows:

We can also write 8y² + 6xy – 3x² as -3x² + 6xy + 8y²

Therefore,

The result is as follows: -9x² + 10xy + 3y²

The subtraction is as follows:

The result is as follows: -14a²b²c + 4ab²c² + 7a²bc²-2a²b²c²

The subtraction is as follows:

The result is as follows: -7t⁴ + 12t³-6t² + 4t + 5

The subtraction is as follows:

The result is as follows: 3ab -7bc + 5ac + 10abc

The subtraction is as follows:

Here 8p(2p-7q) = 16p²-56pq and

7p(3q + 7p) = 21pq + 49p²

Therefore,

The result is as follows: -33p²-77pq

The subtraction is as follows:

Here,3p(-p-a-r) = -3p²-3pa-3pr

Therefore,

The result is as follows: 0p²-3pa-3pr-3pq-3px or -3pa-3pr-3pq-3px

Or -3p(a + r + q + x).

The multiplication is as follows:

(-7pq²r³) ×(-13p3q²r)

= (-7×-13×3)×(pq²r³.pq²r)(Here dot implies multiplication)

= (273)×(p²q⁴r⁴)

= 273p⁴q⁴r⁴

The product is = 273p⁴q⁴r⁴

The multiplication is as follows:

(3x²y²z²)×(17xyz)

= (3×17)×(x² y²z².xyz)

= 51x³y³z³

The product is = 51x³y³z³

The multiplication is as follows:

(15xy²)×(17yz²)

= (15×17)×(xy².yz²)

= 255xy³z²

The product is = 255xy³z²

The multiplication is as follows:

(-5a²bc)×(11ab)×(12abc²)

= (-5×11×12)(a²bc.ab.abc²)

= -660a⁴b³c³

The product is = 255xy³z²

The multiplication is as follows:

(-3x²y)×(5y-xy)

= (-3x²y.5y) + (3x²y.xy)

= -15x²y² + 3x³y²

The product is = -15x²y² + 3x³y²

The multiplication is as follows:

(7prq)×(p-q + r)

= (7prq.p-7prq.q + 7prq.r)

= 7p²rq-7prq² + 7pr²q

The product is = 7p²rq-7prq² + 7pr²q

The multiplication is as follows:

(x²y²z²)(xy-yz + zx)

= (x²y²z².xy-x²y²z².yz + x²y²z².zx)

= x³y³z²-x²y³z³ + x³y²z³

The product is = x³y³z²-x²y³z³ + x³y²z³

The multiplication is as follows:

(p + 6)×(q-7)

= (p.q-7p + 6q-42)

= pq-7p + 6q-42

The product is = pq-7p + 6q-42

The multiplication is as follows:

6mn×0mn

= (6×0)mn = 0

The product is = 0

The multiplication is as follows:

a×a⁵×a⁶

= a¹²

The product is = a¹²

The multiplication is as follows:

b³×3b²×7ab⁵

= 21ab¹⁰

The product is = 21ab¹⁰

The multiplication is as follows:

-7st×-1×-12st²

= -84s²t³

The product is = -84s²t³

The multiplication is as follows:

= ()

=

The product is =

The multiplication is as follows:

(a²-b²)×(a² + b²)

= (a².a² + a².b²-b².a²-b².b²)

= a⁴ + a²b²-a²b²-b⁴

= a⁴ + 0-b⁴

= a⁴-b⁴

The product is = a⁴-b⁴

The multiplication is as follows:

(ab + c)×(ab + c)

= (ab.ab + ab.c + ab.c + c.c)

= a²b² + 2abc + c²

The product is = a²b² + 2abc + c²

The multiplication is as follows:

(pq-2r)×(pq-2r)

= (pq.pq-2pq.r-2pq.r + 2r.2r)

= p²q²-4pqr + 2r²

The product is = p²q²-4pqr + 2r²

The multiplication is as follows:

=

=

= (Here we have calculated the LCM )

=

The product is =

The multiplication is as follows:

=

=

=

=

The product is =

The multiplication is as follows:

(x²-5x + 6)×(2x + 7)

= (x².2x + 7x²-5x.2x-5x.7 + 6.2x + 6.7)

= 2x³ + 7x²-10x²-35x + 12x + 42

= 2x³-3x²-23x + 42

The product is = 2x³-3x²-23x + 42

The multiplication is as follows:

(3x² + 5x-8)×(2x²-4x + 3)

= 3x².2x²-4x.3x² + 3x².3 + 5x.2x²-4x.5x + 5x.3-8.2x² + 8.4x-8.3

= 6x⁴-12x³ + 9x³ + 10x³-20x² + 15x-16x² + 32x-24

= 6x⁴ + 7x³-36x² + 47x-24

The product is = 6x⁴ + 7x³-36x² + 47x-24

The multiplication is as follows:

(2x-2y-3)×(x + y + 5)

= 2x.x + 2x.y + 2x.5-2y.x-2y.y-2y.5-3.x-3.y-3.5

= 2x² + 2xy + 10x-2xy-2y²-10y-3x-3y-15

= 2x²-2y² + 7x-7y-15

The product is = 2x²-2y² + 7x-7y-15

The given expression is (3x + 2y)² + (3x – 2y)²

We have,

(3x + 2y)² = (3x)² + 2.3x.2y + (2y)²(Here we apply standard identities)

= 9x² + 12xy + 4y²

and

(3x-2y)² = (3x)²-2.3x.2y + (2y)²

= 9x²-12xy + 4y²

Therefore,

(3x + 2y)² + (3x – 2y)² = 9x² + 12xy + 4y² + 9x²-12xy + 4y²

= 18x² + 8y²

= 2(9x² + 4y²)

The given expression is (3x + 2y)² - (3x – 2y)²

We have,

(3x + 2y)² = (3x)² + 2.3x.2y + (2y)²

= 9x² + 12xy + 4y²

and

(3x-2y)² = (3x)²-2.3x.2y + (2y)²

= 9x²-12xy + 4y²

Therefore,

(3x + 2y)² - (3x – 2y)² = 9x² + 12xy + 4y²-9x² + 12xy-4y²

= 24xy

The given expression is -ab

= -ab

=

Therefore, - ab =

The given expression is

+ 2xy

= + 2xy

= + 2xy

=

Therefore, + 2xy =

The given expression is

(1.5p + 1.2q)² – (1.5p - 1.2q)²

We have

(1.5p + 1.2q)² = (1.5p)² + 2(1.5p)(1.2q) + (1.2q)²

and

(1.5p - 1.2q)² = (1.5p)²-2(1.5p)(1.2q) + (1.2q)²

Therefore,

(1.5p + 1.2q)² – (1.5p - 1.2q)²

= (1.5p)² + 2(1.5p)(1.2q) + (1.2q)²-(1.5p)² + 2(1.5p)(1.2q)-(1.2q)²

= 2(1.5p)(1.2q)

The given expression is

(2.5m + 1.5q)² + (2.5m - 1.5q)²

We have

(2.5m + 1.5q)² = (2.5m)² + 2(2.5m)(1.5q) + (1.5q)²

(2.5m - 1.5q)² = (2.5m)²-2(2.5m)(1.5q) + (1.5q)²

Therefore,

(2.5m + 1.5q)² + (2.5m - 1.5q)²

= (2.5m)² + 2(2.5m)(1.5q) + (1.5q)² + (2.5m)²-2(2.5m)(1.5q) + (1.5q)²

= 12.50m² + 4.50q²

The given expression is

(x² - 4) + (x² + 4) + 16

= 2x² + 16

Therefore, (x² - 4) + (x² + 4) + 16 = 2x² + 16

The given expression is

(ab - c)² + 2abc

= (ab)²-2(ab)(c) + (c)² + 2abc

= a²b²-2abc + c² + 2abc

= a²b² + c²

Therefore, (ab - c)² + 2abc = a²b² + c²

The given expression is

(a - b) (a² + b² + ab) -(a + b) (a² + b² - ab)

= (a.a² + a.b² + a.ab-b.a²-b.b²-b.ab)-(a.a² + a.b²-a.ab + b.a² + b.b²-b.ab)

= a³ + ab² + a²b-a²b-b³-ab²-a³-ab² + a²b-a²b-b³ + ab²

= -2b³

Therefore, (a - b) (a² + b² + ab) - (a + b) (a² + b² - ab) = -2b³
Method 2:
We know that,
(a - b) (a² + b² + ab) = a³ - b³ and (a + b) (a² + b² - ab) = a³ + b³
Therefore,
(a - b) (a² + b² + ab) - (a + b) (a² + b² - ab) = a³ - b³ - (a³ + b³ ) = -2b³

The given expression is

(b²-49)(b + 7) + 343

= b².b + 7b²-49b-49.7 + 343

= b³ + 7b²-49b-343 + 343

= b³ + 7b²-49b

Therefore, (b²-49)(b + 7) + 343 = b³ + 7b²-49b

The given expression is

(4.5a + 1.5b)² + (4.5b + 1.5a)²

We have,

(4.5a + 1.5b)² = (4.5a)² + 2(4.5a)(1.5b) + (1.5b)²

and

(4.5b + 1.5a)² = (4.5b)² + 2(4.5b)(1.5a) + (1.5a)²

Therefore,

(4.5a + 1.5b)² + (4.5b + 1.5a)²

= (4.5a)² + 2(4.5a)(1.5b) + (1.5b)² + (4.5b)² + 2(4.5b)(1.5a) + (1.5a)²

= (4.5)²(a² + b²) + 2.2(4.5)(1.5)ab + (1.5)²(a² + b²)

= (a² + b²)( (4.5)² + (1.5)²) + 27ab

= 22.5(a² + b²) + 27ab

The given expression is

(pq-qr)² + 4pq²r

= (pq)²-2(pq)(qr) + (qr)² + 4pq²r

= p²q²-pq²r + q²r² + 4pq²r

= p²q² + 3pq²r + q²r²

Therefore , (pq-qr)² + 4pq²r = p²q² + 3pq²r + q²r²

The given expression is

(s²t + tq²)² – (2stq)²

= (s²t)² + 2(s²t)(tq²) + 2(s²t)(tq²)-4s²t²q²

= s⁴t² + 2s²t²q² + 2s²t²q²-4s²t²q²

= s⁴t²

Therefore ,(s²t + tq²)² – (2stq)² = s⁴t²