Question

Get the factorisation ${(a+b+c)}^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)$ writing the expression
${(a+b+c)}^3-a^3-b^3-c^3=[{(a+b+c)}^3-a^3]-[b^3+c^3]$

Answer 1

${(a+b+c)}^3-a^3-b^3-c^3$

$=[{(a+b+c)}^3-a^3]-[b^3+c^3]$

$=\left\{\right(a+b+c-a\left)\right[(a+b+c{)}^2+a^2+(a+b+c\left)a\right]\}-[(b+c)(b^2+c^2-bc)]$

$=\left\{\right(b+c\left)\right[(a+b+c{)}^2+a^2+(a+b+c\left)a\right]\}-[(b+c)(b^2+c^2-bc)]$

$=(b+c)\left[\right(a+b+c{)}^2+a^2+a^2+ab+ac-b^2-c^2+bc]$ $=(b+c)\left[\right(a^2+b^2+c^2+2ab+2bc+2ac)+2a^2+ab+ac-b^2-c^2+bc]$

$=(b+c)[3a^2+3ab+3bc+3ac]$

$=3(b+c)[a^2+ab+bc+ac]$

$=3(b+c)\left[a\right(a+b)+c(b+a\left)\right]$

$=3(b+c)(a+b)(a+c)$

Hence, ${(a+b+c)}^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)$