Question

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$, then $\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}$is equal to .

• A. $\frac{3xyz}{abc}$
• B. $\frac{8xyz}{abc}$
• C. $\frac{9xyz}{abc}$

Answer 1

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

Let $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$

Then, $x=ak,y=bk,z=ck$

LHS: $\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}$

$=\frac{(ak{)}^3}{a^3}+\frac{(bk{)}^3}{b^3}+\frac{(ck{)}^3}{c^3}$

$=3k^3$

Consider the RHS of the options, all of them have $\frac{xyz}{abc}$ in common.

So, $\left(\frac{x}{a}\right)\left(\frac{y}{b}\right)\left(\frac{z}{c}\right)$
= $k.k.k=k^3$

LHS = $k^3$, so we need to multiply 3 to $\left(\frac{x}{a}\right)\left(\frac{y}{b}\right)\left(\frac{z}{c}\right)$ to get $3k^3$

$\therefore \frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=3\times \left(\frac{xyz}{abc}\right)$
Option B, RHS = $3\left(\frac{xyz}{abc}\right)=3k^3=LHS$