Algebraic Identities

The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are used for the factorization of polynomials. Algebraic identities are used in the computation of algebraic expressions and solving different polynomials.

Algebraic Identities

Some Standard Algebraic Identities list are given below:

Identity I

$(a+b)^{2}=a^{2}+2ab+b^{2}$

Example 1: Evaluate $(x+3)(x+3)$ using the identities

$(x+3)(x+3)=x^{2}+2(x)(3)+3^{2}$

$=x^{2}+6x+9$

Example 2: Factorise $49a^{2}+70ab+25b^{2}$

$49a^{2}=(7a)^{2}$ , $25b^{2}=(5b)^{2}$ , $70ab=2(7)(5)ab$

$(7a)^{2}+2(7)(5)ab+(5b)^{2}$

using the identity I $a=7a,b=5b$

$49a^{2}+70ab+25b^{2}=(7a+5b)^{2}=(7a+5b)(7a+5b)$

Identity II

$(a-b)^{2}=a^{2}-2ab+b^{2}$

Example 1: Evaluate $(x-3)(x-3)$ using the identities

$(x-3)(x-3)=x^{2}-2(x)(3)+3^{2}$

$=x^{2}-6x+9$

Identity III

$a^{2}-b^{2}=(a+b)(a-b)$

Example: ${\frac {49}{4}}x^{2}-{\frac {y^{2}}{9}}$

${\frac {49}{4}}x^{2}-{\frac {y^{2}}{9}}$ = $\left[{\frac {7x}{2}}\right]^{2}-\left[{\frac {y}{3}}\right]^{2}$

Using the identity III

${\frac {49}{4}}x^{2}-{\frac {y^{2}}{9}}=$$\left[{\frac {7x}{2}}+{\frac {y}{3}}\right]\left[{\frac {7x}{2}}-{\frac {y}{3}}\right]$

Identity IV

$(x+a)(x+b)=x^{2}+(a+b)x+ab$

Example: Evaluate $103\times 105$ using the identities

Using the identity IV

$103\times 105=(100+3)(100+5)=(100)^{2}+(3+5)(100)+(3\times 5)$

$=10000+800+15=10815$

Identity V

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

Example: Expand $(3a+4b+5c)^{2}$

Using the identity V

$(3a+4b+5c)^{2}=(3a)^{2}+(4b)^{2}+(5c)^{2}+2(3a)(4b)+2(4b)(5c)+2(5c)(3a)$

$(3a+4b+5c)^{2}=9a^{2}+16b^{2}+25c^{2}+24ab+40bc+30ca$

Identity VI

$(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

Example: Evaluate $(104)^{3}$ using the identities

Using the identity VI

$(104)^{3}=(100+4)^{3}=(100)^{3}+(4)^{3}+3(100)(4)(100+4)$

=$1000000+64+124800=1124864$

Identity VII

$(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)$

Example: Evaluate $(99)^{3}$ using the identities

Using the identity VII

$(99)^{3}=(100-1)^{3}=(100)^{3}-(1)^{3}-3(100)(1)(100-1)$

=$1000000-1-29700=970299$

Identity VIII

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Example: Factorise $8x^{3}+y^{3}+27z^{3}-18xyz$

Using the identity VIII

$8x^{3}+y^{3}+27z^{3}-18xyz$

$(2x)^{3}+y^{3}+(3z)^{3}-3(2x)(y)(3z)$

$(2x+y+3z)((2x)^{2}+y^{2}+(3z)^{2}-(2x)(y)-(y)(3z)-(3z)(2x))$

$(2x+y+3z)(4x^{2}+y^{2}+9z^{2}-2xy-3yz-6zx)$