If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}$

Question

What I've done so far; $$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\\ a=bk, b=ck, c=dk\\ a=ck^2, b=dk^2\\ a=dk^3$$

I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$ but couldn't.

Answer 1

Having $a=dk^3,b=dk^2,c=dk$ gives you$$\begin{align}\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}&=\sqrt{\frac{d^5k^{15}+d^2k^4\cdot d^2k^2+d^3k^9\cdot d^2k^2}{d^4k^8\cdot dk+d^4+d^2k^4\cdot dk\cdot d^2}}\\&=\sqrt{\frac{dk^{15}+k^6+dk^{11}}{dk^9+1+dk^5}}\\&=\sqrt{\frac{k^6(dk^9+1+dk^5)}{dk^9+1+dk^5}}\\&=\sqrt{k^6}\\&=|k^3|\end{align}$$

Answer 2

You did fine. Repalicng your values in the term inside the radical, you have $$a^5+b^2c^2+a^3c^2=d^5 k^{15}+d^5 k^{11}+d^4 k^6$$ $$b^4c+d^4+b^2cd^2=d^5 k^9+d^5 k^5+d^4$$ thet is to say that the ratio is $k^6$, its square root is $k^3$ which is $\frac ad$.