WORKSHEET ON ALGEBRAIC IDENTITIES

1)  Expand :

(5x + 3)

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

3)  Find the value : 

[√2 + 1/√ 2]2

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

a2 + b2 + c2

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

8x3 + 27y3

6)  Factor : 

27x3 + 64y3

Detailed Answer Key

1. Answer :

(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

2. Answer : 

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

Add 2ab to each side. 

9 + 2ab  =  29

Subtract 9 from each side.

2ab  =  20

Divide each side by 2. 

ab  =  20

3. Answer :

[√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

4. Answer :

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  =  ab+ 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

5. Answer : 

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

6. Answer :

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)

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