# WORKSHEET ON ALGEBRAIC IDENTITIES

Problem 1 :

Expand :

(5x + 3)

Problem 2 :

If a - b = 3 and a2 + b2 = 29, find the value of ab.

Problem 3 :

Find the value :

[√2 + 1/√ 2]2

Problem 4 :

If (x + a)(x + b)(x + c) = x3 - 10x2 + 45 x - 15, then find the value

a2 + b2 + c2

Problem 5 :

If 2x + 3y = 13 and xy = 6, then find the value of

8x3 + 27y3

Problem 6 :

Factor :

27x3 + 64y3

Problem 7 :

If 2x + 2/x = 3, what is the value of x2 + 1/x2 ?

Problem 8 :

If a + b + c = 6 and a+ b+ c2 = 14, what is the value of

(a - b)2 + (b - c)2 + (c - a)2 ?

Problem 9 :

If x - y = 8 and xy = 5, what is the value of

x- y+ 8(x + y)2

Problem 10 :

If x + y = 5 and xy = 6 and x > y, then find 2(x+ y2). (5x + 3)is in the form of (a + b)2.

(a + b)2  =  a2 + 2ab + b2

Substitute a = 5x and b = 3.

(5x + 3)2  =  (5x)2 + 2 ⋅ 5x ⋅ 3 + 32

=  25x2 + 30x + 9

Given a - b = 3 and a2 + b2 = 29.

To find the value of ab, we can use the following algebraic identity.

(a - b)2  =  a2 - 2ab + b2

or

(a - b)2  =  a2 + b2 - 2ab

Substitute a - b  =  3 and a2 + b2  =  29.

32  =  29 - 2ab

9  =  29 - 2ab

9 + 2ab  =  29

Subtract 9 from each side.

2ab  =  20

Divide each side by 2.

ab  =  20

[√2 + 1/√ 2]2 is in the form of (a + b)2.

(a + b)2  =  a2 + 2ab + b2

Substitute a  =  √2 and b  =  1/√2.

[√2 + 1/√ 2]2  =  (√2)2 + 2 ⋅ √2 ⋅ 1/√2 + (1/√2)2

2 + 2 ⋅ 1 +  1/2

=  2 + 2 +  1/2

=  4 + 1/2

=  9/2

Given : (x + a)(x + b)(x + c)  =  x3 - 10x2 + 45 x - 15

x+ (a+b+c)x+ (ab+bc+ca)x + abc  =  x3 - 10x2 + 45x - 15

Compare the coefficients of x2 and x.

a + b + c  =  - 10

ab + bc + ca  =  45

We can use the following algebraic identity to find the value of (a2 + b2 + c2).

(a + b + c)2  =  a+ b+ c+ 2(ab + bc + ac)

or

a+ b+ c+ 2(ab + bc + ac)  =  (a + b + c)2

Subtract 2(ab + bc + ac) from each side.

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

Substitute.

a+ b+ c =  (-10)- 2(45)

=  100 - 90

=  10

Given :

2x + 3y  =  13

xy  =  6

To find the value of (8x3 + 27y3), write (8x3 + 27y3) as follows.

8x3 + 27y =  (2x)3 + (3y)3

(2x)3 + (3y)3 is in the form of a3 + b3.

a3 + b3  =  (a + b)3 - 3ab(a + b)

Substitute a = 2x and b = 3y.

(2x)3 + (3y)3  =  (2x + 3y)3 - 3 ⋅ 2x ⋅ 3y(2x + 3y)

8x3 + 27y3  =  (2x + 3y)3 - 18xy(2x + 3y)

Substitute 2x + 3y  =  13 and xy  =  6.

=  (13)3 - 18 ⋅ 6 ⋅ (13)

=  2197 - 1404

=  793

We can write 27x3 + 64yas follows.

27x3 + 64y =  (3x)3 + (4y)3

To factor 27x3 + 64y3, we can use the following algebraic identity.

a3 + b3  =  (a + b)(a2 - ab + b2)

Substitute a  =  3x and b  =  4y.

(3x)3 + (4y)3  =  (3x + 4y)[(3x)2 + 3x ⋅ 4y + (4y)2]

27x3 + 64y3  =  (3x + 4y)(9x2 + 12xy + 16y2)

2x + 2/x = 3

Square both sides.

(2x + 2/x)2 = 32

(2x + 2/x)(2x + 2/x) = 9

(2x)2 + (2x)(2/x) + (2/x)(2x) + (2/x)2 = 9

4x2 + 4 + 4 + 4/x2 = 9

4x2 + 8 + 4/x2 = 9

Subtract 8 from both sides.

4x2 + 4/x2 = 1

Factor.

4(x+ 1/x2) = 1

Divide both sides by 4.

x+ 1/x2 = 1/4

Consider the following algebraic identity.

(a + b + c)2 = a+ b+ c+ 2ab + 2bc + 2ca

(a + b + c)2 = a+ b+ c+ 2(ab + bc + ca)

Substitute a + b + c = 6 and a+ b+ c2 = 14.

(6)2 = 14 + 2(ab + bc + ca)

36 = 14 + 2(ab + bc + ca)

Subtract 14 from both sides.

12 = 2(ab + bc + ca)

Divide both sides by 2.

6 = ab + bc + ca

The value of (a - b)2 + (b - c)2 + (c - a)2 :

= (a - b)2 + (b - c)2 + (c - a)2

a- 2ab + b2 + b2- 2bc + c+ c- 2ca + a2

= a2 + a2 + b2 + b+ c2 + c2 - 2ab - 2bc - 2ca

= 2a2 + 2b+ 2c2- 2ab - 2bc - 2ca

= 2(a+ b+ c2) - 2(ab + bc + ca)

Substitute a+ b+ c2 = 14 and ab + bc + ca = 6.

= 2(14) - 2(6)

= 28 - 12

= 16

Consider the square of a binomial given below.

(x - y)= x+ y- 2xy

Substitute x - y = 8 and xy = 5.

82 = x+ y- 2(5)

64 + 10 = x+ y2

74 = x+ y2

Consider the square of a binomial given below.

(x + y)= x+ y+ 2xy

Substitute x+ y2 = 74 and xy = 5.

(x + y)= 74 + 2(5)

(x + y)= 74 + 10

(x + y)= 84

The value of x- y+ 8(x + y)2 :

= x- y+ 8(x + y)2

Use the identity a3 - b3 = (a - b)(a2 + ab + b2).

= (x - y)(x2 + xy + y2) + 8(x + y)2

Substitute.

= (8)(74 + 5) + 8(84)

=  8(79) + 672

= 632 + 672

= 1304

(x + y)2 = (x + y)(x + y)

(x + y)2 = x2 + xy + xy + y2

(x + y)2 = x+ 2xy + y2

or

x+ 2xy + y(x + y)2

Subtract 2xy from both sides.

x+ y= (x + y)- 2xy

Substitute x + y = 5 and xy = 6.

x+ y= 5- 2(6)

x+ y= 25 - 12

x+ y= 13

Multiply both sides by 2.

2(x+ y2) = 2(13)

2(x+ y2) = 26

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