Question

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$ show that $\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{3xyz}{abc}$

Answer 1

Given: $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$ (say)
$\Rightarrow x=ak,y=bk,z=ck$
To show that $\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{3xyz}{abc}$
$L.H.S.=\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}$
$=\frac{a^3{k}^3}{a^3}+\frac{b^3{k}^3}{b^3}+\frac{c^3{k}^3}{c^3}$
$=k^3+k^3+k^3$
$=3k^3$
$=3k\cdot k\cdot k$
$=3\frac{x}{a}\frac{y}{b}\frac{z}{c}(\because x=ak,y=bk,z=ck)$
$=\frac{3xyz}{abc}=$ R.H.S