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

I don't want a full solution, just a little hint on how to start solving are welcome
 [/INST] Note
$$\frac{d^2}{a^2}\cdot {a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}
=\frac{a^3+\frac{b^2c^2}{a^2}+ac^2}{\frac{b^4c}{d^2}+d^2+b^2c}
=\frac{a^3+d^2+ac^2}{a^3+d^2+ac^2}=1$$
Thus,
$$\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}=\sqrt{\frac{a^2}{d^2}}=\frac ad$$
</s>