Question

8 Algebraic identities and their applications ( 2 example each )

Answer 1

Standard Algebraic Identities List

All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:

(a+b)n=nC0.an.b0+nC1.an−1.b1+……..+nCn−1.a1.bn−1+nCn.a0.bn

Some Standard Algebraic Identities list are given below:

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Answer 2

$\huge\star\underline\mathfrak\red{Answer:-}$

Some Standard Algebraic Identities list are given below:

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

 

Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.

Solution: (x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the form Identity I where a = x and b = 1. So we have,

(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1

 

Example 2: Factorise (x4 – 1) using standard algebraic identities.

Solution: (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,

(x4 – 1) = ((x2)2– 12) = (x2 + 1)(x2 – 1)

The factor (x2 – 1) can be further factorised using the same Identity III where a = x and b = 1. So,

(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)

 

Eample 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.

Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,

16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)

 

Example 4: Expand (3x – 4y)3 using standard algebraic identities.

Solution: (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,

(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2

 

Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.

Solution: (x3 + 8y3 + 27z3 – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,

(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)