Question

If ${\displaystyle \frac{a}{b}=\frac{c}{d}=\frac{e}{f}}$, prove that each ratio is ${\displaystyle {\left[\frac{2a^3+5c^3+7e^3}{2b^3+5d^3+7f^3}\right]}^{\frac{1}{3}}}$

Answer 1

${\displaystyle \frac{a}{b}=\frac{c}{d}=\frac{e}{f}}$ = k(say)
∴ a = k, c = dk, e = fk
${\displaystyle {\left[\frac{2a^3+5c^3+7e^3}{2b^3+5d^3+7f^3}\right]}^{\frac{1}{3}}}$

= ${\displaystyle {\left[\frac{2b^3{k}^3+5d^3{k}^3+7f^3{k}^3}{2b^3+5d^3+7f^3}\right]}^{\frac{1}{3}}}$

= ${\displaystyle k{\left[\frac{2b^3+5d^3+7f^3}{2b^3+5d^3+7f^3}\right]}^{\frac{1}{3}}}$ = k
Hence proved.