Question

State True or False.

$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

• A. True
• B. False

Answer 1

The correct option is A True

$R.H.S=\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=\frac{1}{2}(a+b+c\left)\right[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)]$
$=\frac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=a^3+a{b}^2+a{c}^2-a^{2}b-abc-c{a}^2+a^{2}b+b^3+c^{2}b-a{b}^2-b^{2}c-cab+a^{2}c+b^{2}c+c^3-abc-b{c}^2-c^{2}a$
$=a^3+b^3+c^3-3abc$