Question

If $x=a(b-c);y=b(c-a);z=c(a-b)$, then ${\left(\frac{x}{a}\right)}^3+{\left(\frac{y}{b}\right)}^3+{\left(\frac{z}{c}\right)}^3$ is equal to:

• A. $\frac{xyz}{abc}$
• B. $\frac{1}{3}\frac{xyz}{abc}$
• C. $3\frac{xyz}{abc}$
• D. $\frac{3(x+y+z)}{\left(abc\right)}$

Answer 1

The correct option is C $3\frac{xyz}{abc}$

Let $\frac{x}{a}=b-c$, $\frac{y}{b}=c-a$ and $\frac{z}{c}=a-b$.

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

Here, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=b-c+c-a+a-b=0$, therefore,

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

Hence, ${\left(\frac{x}{a}\right)}^3+{\left(\frac{y}{b}\right)}^3+{\left(\frac{z}{c}\right)}^3=3\frac{xyz}{abc}$.

Therefore, option $C$ is correct.