
Simplification
Hello,
If $\displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{d}, $ prove that
$\displaystyle \frac{a}{d} $ is equal to $\displaystyle \sqrt{\frac{a^5 + b^2c^2 + a^3c^2}{b^4c + d^4 + b^2cd^2}} $
It seems like a mess. Any suggestions on where to start?
Btw, this problem is from Higher Algebra by H.S Hall and S. R Knight, a textbook that is still used in India even though it was written in the 1800's.

We are assuming that $\displaystyle a,b,c,d>0$, right? (If not, you could have, say, $\displaystyle a=c=1$ and $\displaystyle b=d=1$ and then $\displaystyle \frac{a}{d}$ would be negative.)
Looking at the expression, the best thing to do first looks like eliminating $\displaystyle b^2=ac$:
$\displaystyle \color{white}.\quad.$ $\displaystyle \frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}$
$\displaystyle =\ \frac{a^5+ac^3+a^3c^2}{a^2c^3+d^4+ac^2d^2}$
We also have $\displaystyle a^2d^2=b^2c^2=ac^3$, i.e. $\displaystyle c^3=ad^2$. So
$\displaystyle \color{white}.\quad.$ $\displaystyle \frac{a^5+ac^3+a^3c^2}{a^2c^3+d^4+ac^2d^2}$
$\displaystyle =\ \frac{a^5+a^2d^2+a^3c^2}{a^3d^2+d^4+ac^2d^2}$
$\displaystyle =\ \frac{a^2\left(a^3+d^2+ac^2\right)}{d^2\left(a^3+d ^2+ac^2\right)}$
$\displaystyle =\ \frac{a^2}{d^2}$