Question

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ prove that:
$\frac{a^3+c^3+e^3}{b^3+d^3+f^3}=\frac{ace}{bdf}$

Answer 1

Given:- $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$

To proof:- $\frac{a^3+c^3+e^3}{b^3+d^3+f^3}=\frac{ace}{bdf}$

Proof:-

Let $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$

$\Rightarrow a=bk,c=dk,e=fk$

Now,

$\frac{a^3+c^3+e^3}{b^3+d^3+f^3}=\frac{ace}{bdf}$

$\frac{{\left(bk\right)}^3+{\left(dk\right)}^3+{\left(fk\right)}^3}{b^3+d^3+f^3}=\frac{\left(bk\right)\left(dk\right)\left(fk\right)}{bdf}$

$\frac{b^3{k}^3+d^3{k}^3+f^3{k}^3}{b^3+d^3+f^3}=\frac{bdf\cdot k^3}{bdf}$

$\frac{k^3(b^3+d^3+f^3)}{b^3+d^3+f^3}=\frac{\left(bdf\right)k^3}{bdf}$

$\Rightarrow k^3=k^3$

L.H.S. $=$ R.H.S.

Hence proved