Question

Evaluate $\frac{a^3+b^3+c^3-3abc}{(a+b+c)},\text{where}(a+b+c)\ne 0$.

• A. $a^2+b^2+c^2-ab-bc-ca$
• B. $a^2+b^2+c^2+ab-bc-ca$
• C. $a^2+b^2+c^2+ab+bc+ca$
• D. $a^2-b^2-c^2+ab+bc+ca$
  

Answer 1

The correct option is A $a^2+b^2+c^2-ab-bc-ca$

We have the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Taking (a+b+c) from R.H.S. to L.H.S., we get

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