Question

$\frac{x^3+y^3+z^3-3xyz}{a^3+b^3+c^3-3abc}\times \frac{a^2+b^2+c^2-ab-bc-ca}{x^2+y^2+z^2-xy-yz-zx}$ equals

• A. $1$
• B. $\frac{x^2+y^2+z^2}{a^2+b^2+c^2}$
• C. $\frac{x+y+z}{a+b+c}$
• D. $\frac{xyz}{abc}$

Answer 1

The correct option is C $\frac{x+y+z}{a+b+c}$

Reqd exp
= $\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}\times \frac{(a^2+b^2+c^2-ab-bc-ca)}{(x^2+y^2+z^2-xy-yz-zx)}=\frac{x+y+z}{a+b+c}$