Question

If $x+y+z=0$ and $\frac{x^3+y^3}{xyz}=\frac{y^3+z^3}{xyz}=\frac{x^3+z^3}{xyz}=a$, then which of the following can be a?

• A. $2$
• B. $3$
• C. $6$
• D. $9$

Answer 1

The correct option is A $2$

Let, $x+y+z=0$.....(1) and $\frac{x^3+y^3}{xyz}=\frac{y^3+z^3}{xyz}=\frac{x^3+z^3}{xyz}=a$........(2).

From (2) each ratio $=\frac{(x^3+y^3)+(y^3+z^3)+(z^3+x^3)}{xyz+xyz+xyz}$

$=\frac{2(x^3+y^3+z^3)}{3xyz}$

$=\frac{2\left(3xyz\right)}{3xyz}=2.$[From (1) $x+y+z=0\Rightarrow (x+y{)}^3=-y^3\Rightarrow x^3+y^3+z^3=3xyz$]

So, $a=2$.