Question

Factorise: (a+b+c)^3-a^3-b^3-c^3.​

Answer 1

The factor is $\Rightarrow 3(a+b)(b+c)(c+a)$

Step-by-step explanation:

Given: $(a+b+c)^3-a^3-b^3-c^3$

Factorize the above expression

Formula:

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

• $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

$\Rightarrow (a+b+c)^3-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)-3abc$

$\Rightarrow (a+b+c)[(a+b+c)^2-(a^2+b^2+c^2-ab-bc-ca)]-3abc$

$\Rightarrow (a+b+c)[a^2+b^2+c^2+2ab+2bc+2ca-a^2-b^2-c^2+ab+bc+ca]-3abc$

$\Rightarrow (a+b+c)[3ab+3bc+3ca]-3abc$

$\Rightarrow 3(a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2)$

$\Rightarrow 3(a+b)(b+c)(c+a)$

#BAL

#ANSWERWITHQUALITY