If $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$, show that :

$\dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} = \dfrac{3xyz}{abc}$.

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

∴ x = ak, y = bk and z = ck.

Substituting x = ak, y = bk and z = ck in L.H.S. of $\dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} = \dfrac{3xyz}{abc}$,

$\Rightarrow \dfrac{(ak)^3}{a^3} + \dfrac{(bk)^3}{b^3} + \dfrac{(ck)^3}{c^3} \\[1em] \Rightarrow \dfrac{a^3k^3}{a^3} + \dfrac{b^3k^3}{b^3} + \dfrac{c^3k^3}{c^3} \\[1em] \Rightarrow k^3 + k^3 + k^3 \\[1em] \Rightarrow 3k^3.$

Substituting x = ak, y = bk and z = ck in R.H.S. of $\dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} = \dfrac{3xyz}{abc}$,

$\Rightarrow \dfrac{3(ak)(bk)(ck)}{abc} \\[1em] \Rightarrow \dfrac{3abck^3}{abc} \\[1em] \Rightarrow 3k^3.$

Since, L.H.S. = R.H.S. = 3k³

Hence, proved that $\dfrac{x^3}{a^3} + \dfrac{y^3}{b^3} + \dfrac{z^3}{c^3} = \dfrac{3xyz}{abc}$.