Evaluate each of the following using identities:
(i)
(ii) (2x+y) (2x-y)
(iii)
(iv) (a – 0.1) (a +0.1)
(v)
(i) We know, (a-b)2 = a2 + b2 -2ab
(ii) We know, (a-b)2 = (a+b) (a-b)
(2x+y) (2x-y) = (2x)2– y2 = 4x2 – y2
(iii) We know, (a-b)2 = a2 + b2 -2ab
(iv) We know, (a-b)2 = (a+b) (a-b)
(a – 0.1) (a +0.1) = (a)2– (0.1)2
= a2 – 0.01
(v) We know, (a-b)2 = a2 - b2
Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991×1009
(iv) 117×83
(i) We will use the identity, (a-b)2 = a2 + b2 – 2ab
(ii) We will use the identity, (a-b)2 = a2 + b2 – 2ab
(iii) We will use the identity, (a-b)(a+b)= a2 - b2
(iv) We will use the identity, (a-b) (a+b)= a2 - b2
Simplify each of the following:
(i) 175 × 175 + 2 × 175 × 25 + 25 ×25
(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22
(iii) 0.76 ×0.76+2×0.76×0.24+0.24×0.24
(iv)
(i) We know, (a+b)2 = a2 + b2 +2ab
(ii) We know, (a-b)2 = a2 + b2 -2ab
(iii) We know, (a+b)2 = a2 + b2 +2ab
(iv) We know, (a-b)2 = (a-b) (a+b)
If x + = 11, find the value of x2 +
Here, we will use (a+b)2 = a2 + b2 +2ab
If x - = -1, find the value of x2 +
Here, we will use (a-b)2 = a2 + b2 -2ab
If x + = , find the value of x2 + and x4+
Here, we will use (a+b)2 = a2 + b2 +2ab
Now,
If x2 + = 66, find the value of x -
Here, we will use (a-b)2 = a2 + b2 -2ab.
If x2 + = 79, find the value of x +
Here, we will use (a+b)2 = a2 + b2 + 2ab
If 9x2 +25y2 = 181 and xy = -6, find the value of 3x +5y
Here, we will use (a+b)2 = a2 + b2 + 2ab
If 2x +3y = 8 and xy = 2, find the value of 4x2 +9y2
Here, we will use (a+b)2 = a2 + b2 + 2ab
If 3x - 7y = 10 and xy = -1, find the value of 9x2 + 49y2
Here, we will use (a+b)2 = a2 + b2 + 2ab
Simplify each of the following products:
(i)
(ii)
(iii) -x2 + 2x
(iv) (x2 + x – 2) (x2 - x + 2)
(v) (x3 - 3x2 – x) (x2 - 3x + 1)
(vi) (2x4 - 4x2 + 1) (2x4 - 4x2 - 1)
(i) On regarranging we get,
(ii) On regarranging we get,
(iii) On rearranging we get, -x2 + 2x
=
Using, (a-b)2 = a2 + b2 – 2ab
(iv) Using the idendity, (a+b)(a-b) = a2-b2
On rearranging we get,
(x2 + x – 2) (x2 - x + 2) = {x2 + (x – 2)} {(x2 – (x - 2)}
= (x2)2 – (x – 2)2 = x4-(x2 - 4x + 4)
= x4 – x2 + 4x – 4
(v) Taking x as common factor, we write,
= x (x2 - 3x – 1) (x2 - 3x + 1)
= {x (x2 - 3x – 1)} (x2 - 3x + 1)
= x [{(x2 - 3x) – 1)} {(x2 - 3x)+1)}]
= x {(x2 - 3x)2 – 12}
= x (x4 - 6x3+9x2-1)
= x5 – 6x4 + 9x3 -x
(vi) On Reaaranging we get,
(2x4 - 4x2 + 1) (2x4 - 4x2 - 1)
= {(2x4 - 4x2) + 1} {(2x4 - 4x2)- 1)}
= (2x4 - 4x2)2 – 12
= 4x8 + 16x4 -2 × 2x4 × 4x2 – 1
= 4x8 + 16x4 -16x6 -1
Prove that a2+ b2+c2–ab–bc–ca is always non-negative for all values of a, b and c.
We have to prove that a2+ b2+c2–ab–bc–ca 0
Lets us consider,
Write the following in the expanded form:
(i) (a + 2b + c)2
(ii) (2a - 3b - c)2
(iii) (-3x + y + z)2
(iv) (m + 2n – 5p)2
(v) (2 + x – 2y)2
(vi) (a2 + b2 + c2)2
(vii) (ab + bc + ca)2
(viii)
(ix)
(x) (x + 2y + 4z)2
(xi) (2x - y + z)2
(xii) (-2x + 3y + 2z)2
(i) Using idendity,
(ii) Using idendity,
(iii) Using idendity,
(iv) Using idendity,
(v) Using idendity,
(vi) Using idendity,
(vii) Using idendity,
(viii) Using idendity,
(ix) Using idendity,
(x) Using idendity,
(xi) Using idendity,
(xii) Using idendity,
Simplify:
(i) (a+b+c)2 +(a-b+c)2
(ii) (a+b+c)2 -(a-b+c)2
(iii) (a+b+c)2 +(a-b+c)2+(a+b-c)2
(iv) (2x+p+c)2-(2x-p+c)2
(v) (x2+y2-z2)2-(x2-y2+z2)2
(i) Using idendity,
(ii) Using idendity,
(iii) Using idendity,
(iv) Using idendity,
(v) Using identity: a2 - b2 = (a + b)(a - b)
(x2 + y2 - z2)2 - (x2 -y2 +z2)2
=(x2 + y2 - z2 +(x2 -y2 +z2)) (x2 + y2 - z2 -(x2 -y2 +z2))
= 2x2(2y2 - 2z2)
= 4x2y2 -4x2z2
If a+b+c =0 and a2+b2+c2 =16, find the value of ab + bc + ca.
Using idendity,
Given: a+b+c = 0 and a2 + b2 + c2 =16
Squaring the equation, a+b+c = 0 on both the sides, we get,
If a2+b2+c2 =16, and ab + bc + ca=10, find the value of a+b+c.
Using the identity,
If a+b+c =9 and ab+bc+ca=23, find the value of a2+b2+c2.
Usint the identity,
Find the value of 4x2+y2+25z2+4xy-10yz-20zx when x = 4, y = 3 and z = 2.
Using the identity,
Simplify each of the following expressions :
(i) (x+y+z)2+
(ii) (x+y-2z)2-x2 – y2 - 3z2+4xy
(iii) (x2 – x + 1)2 - (x2 + x + 1)2
(i) Using identity,
(ii) Using the identity,
(iii) Using the identity,
Find the cube of each of the following binomial expressions:
(i)
(ii)
(iii)
(iv)
(i) Using the identity, (a+b)3 = a3 + b3 + 3a2 b + 3ab2
We will write the bionomial expression,
(ii) Using the identity, (a-b)3 = a3 - b3 - 3a2 b + 3ab2
We will write the bionomial expression,
(iii) Using the identity, (a-b)3 = a3 - b3 - 3a2 b + 3ab2
We will write the bionomial expression,
(iv) Using the identity, (a-b)3 = a3 - b3 - 3a2 b + 3ab2
We will write the bionomial expression,
Simplify each of the following :
(i) (x+3)3 + (x-3)3
(ii)
(iii)
(iv) (2x-5y)3 - (2x+5y)3
(i) Using the identity,
(ii) Using the identity, a3 - b3 = (a-b)(a2 +ab + b2)
(iii) Using the identity, a3 + b3 = (a+b)(a2 - ab + b2)
(iv) Using the identity, a3 - b3 = (a-b)(a2 + ab + b2)
If a+b=10 and ab=21, find the value of a3+b3.
Given (a+b)=10 and ab = 21
Using, (a+b)3 = a3 + b3 + 3ab(a+b), we get,
If a-b=4 and ab=21, find the value of a3-b3.
Given (a - b)=4 and ab = 21
Using, (a-b)3 = a3 - b3 - 3ab(a-b), we get,
If x+=5, find the value of x3+.
Given: x+=5,
Using, (a+b)3 = a3 + b3 + 3ab(a+b), we get
If x-= 7, find the value of x3-.
Given: x-= 7
Using, (a-b)3 = a3 - b3 - 3ab(a-b), we get
If x-=5, find the value of x3-.
Given: x-=5
Using, (a+b)3 = a3 + b3 + 3ab(a+b), we get
If x2+=51, find the value of x3-.
Using the identity, (x+y)2=x2 + y2 + 2xy
If x2+=98, find the value of x3+.
Using the identity, (x+y)2=x2 + y2 + 2xy
If 2x+3y=13 and xy=6, find the value of 8x3+27y3.
Given :- 2x+3y=13 and xy=6.
Using, (a+b)3 = a3 + b3 + 3ab(a+b), we get
If 3x-2y=11 and xy=12, find the value of 27x3-8y3.
Given :- 3x-2y=11 and xy=12.
Using, (a-b)3 = a3 - b3 - 3ab(a-b), we get
If x4+=119, find the value of x3-.
Given :- x4+=119
Using, (a+b)2 = a2 + b2 +2ab, we get
Evaluate each of the following:
(i) (103)3
(ii) (98)3
(iii) (9.9)3
(iv) (10.4)3
(v) (598)3
(vi) (99)3
(i) Using the identity,
(ii) Using the identity,
(iii) Using the identity,
(iv) Using the identity,
(v) Using the identity,
(vi) Using the identity,
Evaluate each of the following :
(i) 1113-893
(ii) 463+343
(iii) 1043+963
(iv) 933-1073
(i) Uisng the identity:
(ii) Uisng the identity:
(iii) Using the identity:
(iv) Uisng the identity:
If x+= 3, Calculate x2 +, x3 +and x4+.
Given x+= 3
If x4+=194, find x3 +, x2 +and x+
Given x4+=194
Find the value of 27x3 + 8y3, if
(i) 3x + 2y = 14 and xy=8
(ii) 3x + 2y = 20 and xy=
(i) Using,
(ii) Using the identity, we get,
Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz=12.
Using the identity, we write,
If x-= 3+2 , find the value of x3 -.
Using the identity, we write,
Find the following products:
(i) (3x+2y)(9x2-6xy+4y2)
(ii) (4x-5y)(16x2+20xy+25y2)
(iii) (7p4+q)(49p8-7p4q+q2)
(iv)
(v)
(vi)
(vii)
(viii)
(ix) (1+x)(1+x+x2)
(x) (1+x)(1-x+x2)
(xi) (x2-1)(x4+x2+1)
(xii) (x3+1)(x6-x3+1)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
If x = 3 and y = -1, find the values of each of the following using in identity:
(i) (9y2 -4x2) (81y4 + 36x2y2 +16x4)
(ii)
(iii)
(iv)
(v)
(i)
(ii)
(iii)
(iv)
(v)
If a+b=10 and ab=16, find the value of a2– ab+b2 and a2+ab+b2.
Given: a+b=10 and ab =16
To find: a2– ab+b2
To find: a2+ ab+b2
If a+b=8 and ab=6, find the value of a3+b3.
Given: a+b=8 and ab =6
To find: a3+b3
If a-b=6 and ab=20, find the value of a3-b3.
Given: a-b=6 and ab =20
To find: a3-b3
If x = -2 and y = 1, by using an identity find the value of the following:
(i) (4y2–9x2) (16y4+36x2y2+81x4)
(ii)
(iii)
(i)
(ii)
(iii)
Find the following product:
(i) (3x+2y+2z) (9x2+4y2+4z2-6xy-4yz-6zx)
(ii) (4x-3y+2z) (16x2+9y2+4z2+12xy+6yz-8zx)
(iii) (2a-3b-2c) (4a2+9b2+4c2+6ab-6bc+4ca)
(iv) (3x-4y+5z)(9x2+16y2+25z2+12xy-15zx+20yz)
(i) Using the identity,
(ii) Using the identity,
(iii) Using the identity,
(iv) Using the identity,
If x+y+z = 8 and xy+yz+zx =20, find the value of x3 +y3 +z3 -3xyz.
In x3 +y3 +z3 -3xyz,
Using the identity
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) we get,
x3 +y3 +z3 -3xyz = ( x + y +z) (x2+y2+z2-xy-yz-zx)
x3 +y3 +z3 -3xyz = ( x + y +z) [x2+y2+z2-(xy+yz+zx)] …. (1)
We also know,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
For x2+y2+z2
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz+ zx)
⇒ 82 = x2 + y2 + z2 + 2(20)
⇒ 64 = x2 + y2 + z2 + 40
⇒ x2 + y2 + z2 = 24
From (1) we get,
x3 +y3 +z3 -3xyz = 8 [24-20]
= 8(4)
= 32
If a+b+c=9 and ab+bc+ca=26, find the value of a3 +b3 +c3 -3abc.
Using the identity,
If a+b+c=9 and a2+b2+c2=35, find the value of a3 +b3 +c3 -3abc.
Using the identity,
Evaluate :
(i) 253 – 753 + 503
(ii) 483 – 303 - 183
(iii)
(iv) (0.2)3 – (0.3)3 + (0.1)3
(i)
(ii)
(iii)
(iv)
If x+ =3, then find the value of x2+.
Given: x+ =3
If x+ =5, then x2+=
A. 25
B. 10
C. 23
D. 27
If x+ =3, then find the value of x6+.
We are given that x+ =3
If x+ =2, then x3+=
A. 64
B. 14
C. 8
D. 2
If a+b=7 and ab=12, find the value of a2+b2.
Using the identity,
If x+ =4, then x4+ =
A. 196
B. 194
C. 192
D. 190
If a-b=5 and ab=12, find the value of a2+b2.
Using the identity,
If x+ =3, then x6+ =
A. 927
B. 414
C. 364
D. 322
If x- =, then write the value of 4x2+.
Given: x- =
If x4+ = 623, then x+=
A. 27
B. 25
C. 3
B. -3
If x2+= 102, then x-=
A. 8
B. 10
C. 12
D. 13
If a2+ =102, find the value of a-.
Here, we will use (a+b)2 = a2 + b2 + 2ab
If a+b+c=0 then write the value of.
If x3+=110, then x +=?
A. 5
B. 10
C. 15
D. none of these
If x3-=14, then x -=
A. 5
B. 4
C. 3
D. 2
If x4+ = 194, then x3-=
A. 76
B. 52
C. 64
D. none of these
If x- = , then x +=
A. 4
B.
C.
D.
If 3x+ =7, then =
A. 25
B. 35
C. 49
D. 30
If a2+b2 + c2-ab-bc-ca =0, then
A. a + b =c
B. b + c = a
C. c + a = b
D. a = b = c
If a + b + c =0, then, =
A. 0
B. 1
C. -1
D. 3
If a1/3+b1/3 + c1/3 =0, then
A. a+b+ c =0
B. (a+b+ c)3 =27abc
C. a+b+ c =3abc
D. a3+b3+ c3 =0
If a+b+ c =9 and ab + bc + ca = 23, then a2+b2 + c2=
A. 35
B. 58
B. 127
D. none of these
Hence, 92 = a2 + b2 + c2 + 2 × 23
⇒ a2 + b2 + c2 = 35
If a+b+ c =9, then ab+bc+ca=23, then a3+b3 + c3 – 3abc =
A. 108
B. 207
C. 669
D. 729
(a-b)3+(b-c)3+ (c-a)3=
A. (a+b+ c) (a2+b2 + c2-ab-bc-ca)
B. (a-b)(b-c) (c-a)
C. 3 (a-b)(b-c) (c-a)
D. none of these
Solve the equation and choose the correct answer:
=
A. 3(a+b)(b+c) (c+a)
B. 3 (a-b)(b-c) (c-a)
C. (a-b)(b-c) (c-a)
D. None of these
The product (a+b) (a-b) (a2-ab+b2) (a2+ab+b2) is equal to
A. a6+b6
B. a6-b6
C. a3-b3
D. a3+b3
If = -1, then a3-b3 =
A. 1
B. -1
C.
D. 0
The product (x2-1) (x4+x2+1) is equal to=
A. x8-1
B. x8+1
C. x6-1
D. x6+1
If a-b=-8, and ab =-12, then a3 – b3 =
A. -244
B. -240
C. -224
D. -260
If the volume of a cuboid is 3x2-27, then its possible dimensions are
A. 3, x2, -27x
B. 3, x-3, x+3
C. 3, x2, 27x
D. 3, 3, 3
Thus, the possible dimensions are 3, (x+3)(x-3)
If = 1, then a3+b3 =
A. 1
B. -1
C.
D. 0
75×75+2×75×25+25×25 equal to
A. 10000
B. 6250
C. 7500
D. 3750
We know, (a+b)2 = a2 + b2 +2ab
(x-y)(x+y)(x2+y2)(x4+y4) is equal to
A. x16 – y16
B. x8 –y8
C. x8+y8
D. x16 + y16
If 48a2-b= , then the value of b is
A. 0
B.
C.
D.