Question

If $x=a+b;y=b+c;z=c+a$, then the value of $\frac{x^3+y^3+z^3-3xyz}{a^3+b^3+c^3-3abc}$ is equal to:

• A. $3$
• B. $2$
• C. $1$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $2$

$x^3+y^3+z^3-3xyz$

=$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$=(x+y+z)\frac{1}{2}[{(x-y)}^2+{(y-z)}^2+{(z-x)}^2]$ ....(i).

Now, $x+y+z=a+b+b+c+c+a=2(a+b+c)$ and

$x-y=a-c;y-z=b-a$ and $z-x=c-b$.

$\therefore$ (i) can be written as $2(a+b+c).\frac{1}{2}[{(a-c)}^2+{(b-a)}^2+{(c-b)}^2]$

$=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=2(a^3+b^3+c^3-3abc)$.

$\therefore$ The value of the given expression $\Rightarrow \frac{x^3+y^3+z^3-3xyz}{a^3+b^3+c^3-3abc}=2$.

Therefore, option $B$ is correct.