<s>[INST] If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}$ 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.
 [/INST] 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 $\frac ad$.
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