Algebraic Identities – Solved examples

Algebraic identities

The algebraic identities are specific types of algebraic expressions which are valid for any values of the variables present in them. These can be used for finding the values of variables, expansion or factorizing polynomials.

There are some standard identities as given below. All other identities can be derived from these standard identities.

Standard Identities

1. (a+b)^{2}=a^{2}+b^{2}+2 a b
2. (a-b)^{2}=a^{2}+b^{2}-2 a b
3. (a+b)(a-b)=a^{2}-b^{2}
4. (x+a)(x+b)=x^{2}+(a+b) x+a b
5. (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a
6. (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)
7. (a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)
8. a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c

Proof of some basic identities (with the help of geometry)

We are going to prove the 3 most basic identities with the help of geometry.

\text { 1. }(a+b)^{2}=a^{2}+b^{2}+2 a b

In the figure given above the square having side ‘a + b’ is divided into four parts.

Area of the square having sides (a+b)=(a+b)=a^{2}+b^{2}+a b+a b

\Rightarrow(a+b)^{2}=a^{2}+b^{2}+2 a b

\text { 2. }(a-b)^{2}=a^{2}+b^{2}-2 a b

In the figure given above the square having side ‘a’ is divided into three parts

Area of the square having side ‘a’ =a^{2}=(a-b)^{2}+b(a-b)+a b

\Rightarrow a^{2}=(a-b)^{2}+a b-b^{2}+a b
\Rightarrow a^{2}=(a-b)^{2}-b^{2}+2 a b
\Rightarrow(a-b)^{2}=a^{2}+b^{2}-2 a b

In the figure given above the rectangle having length and breadth ‘x+a’ and ‘x+b’ respectively are divided into four parts

Area of rectangle having sides (x+a) and (x+b)=(x+a)(x+b)=x^{2}+x a+x b+a b

\Rightarrow \quad(x+a)(x+b)=x^{2}+(a+b) x+a b

Solved Examples

Solution

\text { First, we have to write }(a+b)^{3} \text { as a product of }(a+b) \text { and }(a+b)^{2}

\text { LHS }=(a+b)^{3}=(a+b)(a+b)^{2}
=(a+b)\left(a^{2}+b^{2}+2 a b\right)
=a\left(a^{2}+b^{2}+2 a b\right)+b\left(a^{2}+b^{2}+2 a b\right)
=a^{3}+a b^{2}+2 a^{2} b+b a^{2}+b^{3}+2 a b^{2}
=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}
=a^{3}+b^{3}+3 a b(a+b)=\text { RHS }=\text { Hence proved }

\text { 2) Prove that }(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a

Solution:

Let, a+b=x
\text { LHS }=(a+b+c)^{2}=(x+c)^{2}
=x^{2}+c^{2}+2 x c
=(a+b)^{2}+c^{2}+2(a+b) c
=a^{2}+b^{2}+2 a b+c^{2}+2 c a+2 b c
=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a= RHS = Hence proved

\text { 3) Prove that }(a+b+c)^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c

Solution

(a+b+c)^{3}=(x+c)^{3}=x^{3}+c^{3}+3 x c(x+c)
=x^{3}+c^{3}+3 x^{2} c+3 x c^{2}
=(a+b)^{3}+c^{3}+3(a+b)^{2} c+3(a+b) c^{2}
=a^{3}+b^{3}+3 a b(a+b)+c^{3}+3\left(a^{2}+b^{2}+2 a b\right) c+3(a+b) c^{2}
=a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a b^{2}+3 a^{2} c+3 b^{2} c+6 a b c+3 a c^{2}+3 b c^{2}
=a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a b^{2}+3 a b c+3 b^{2} c+3 b c^{2}+3 a b c+3 a^{2} c+3 a b c+3 a c^{2}-3 a b c
=a^{3}+b^{3}+c^{3}+3 a b(a+b+c)+3 b c(a+b+c)+3 c a(a+b+c)-3 a b c
=a^{3}+b^{3}+c^{3}+3(a+b+c)(a b+b c+c a)-3 a b c