Question

If $b$ is the mean proportion between $a$ and $c$, show that.
$\frac{a^4+a^2{b}^2+b^4}{b^4+b^2{c}^2+c^4}=\frac{a^2}{c^2}$

Answer 1

We have

$\frac{a^4+a^2{b}^2+x^4}{b^4+b^2{c}^2+c^4}$ $=\frac{a^2}{c^2}$

We know that,

Mean proportion, $b^2=ac$

$\therefore \frac{a^4+a^{2}ac+(a^2{c}^2{)}^2}{(a^2{c}^2{)}^2+ac\cdot c^2+c^4}=\frac{a^2}{c^2}$

Taking L.H.S only

$\Rightarrow \frac{a^4+a^{3}c+a^4{c}^4}{c^4+c^{3}a+a^4{c}^4}=\frac{a^2}{c^2}$

$\Rightarrow \frac{a^3(a+c+a{c}^4)}{c^3(c+a+c{a}^4)}=\frac{a^2}{c^2}$

$\Rightarrow \frac{a^2+ac+a^2{c}^4}{c^2+ac+c^2{a}^4}=1$

$\Rightarrow a^2+ac+a^2{c}^4=c^2+ac+c^2{a}^4$

$\Rightarrow a^2-c^2=c^2{a}^4-a^2{c}^4$

$\Rightarrow (a^2-c^2)=a^2{c}^2(a^2-c^2)$

$\Rightarrow \frac{a^2}{c^2}=$R.H.S