Question

Question 9
Prove that $(a+b+c{)}^3–a^3–b^3–c^3=3(a+b\left)\right(b+c\left)\right(c+a)$.

Answer 1

To prove, $(a+b+c{)}^3–a^3–b^3–c^3=3(a+b\left)\right(b+c\left)\right(c+a)$.
$LHS=\left[\right(a+b+c{)}^3–a^3]–(b^3+c^3)$
$=(a+b+c-a)\left[\right(a+b+c{)}^2+a^2+a(a+b+c)]-[(b+c)(b^2+c^2-bc)]$
[Using identity, $a^3+b^3=(a+b)(a^2+b^2-ab)$ and $a^3–b^3=(a-b)(a^2+b^2+ab)$]
$=(b+c)[a^2+b^2+c^2+2ab+2bc+2ca+a^2+a^2+ab+ac]-(b+c)(b^2+c^2-bc)$
$=(b+c)[b^2+c^2+3a^2+3ab+3ac-b^2-c^2+3bc]$
$=(b+c)\left[3\right(a^2+ab+ac+bc\left)\right]$
$=3(b+c)\left[a\right(a+b)+c(a+b\left)\right]$
$=3(b+c)\left[\right(a+c\left)\right(a+b\left)\right]$
$=3(a+b)(b+c)(c+a)=RHS$
Hence proved.