Question

The value of $\frac{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}{(x-y)(y-z)(z-x)}$ is equal to:

• A. $4$
• B. $1$
• C. $3$
• D. $2$

Answer 1

The correct option is C $3$

Let $a=(x-y)$, $b=(y-z)$ and $c=(z-x)$.

We know that $a^3+b^3+c^3=3abc$, when $a+b+c=0$.

Since $a+b+c=x-y+y-z+z-x=0$, therefore $a^3+b^3+c^3=3abc$.

Now consider,

$\frac{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}{(x-y)(y-z)(z-x)}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3$.

Hence, $\frac{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}{(x-y)(y-z)(z-x)}=3$.

Therefore, option $C$ is correct.