Prove the equation without solving for X My niece asked me this - If $x=1/(5-x)$ prove $x^3 + \dfrac{1}{x^3}=110$ without solving for x. I said its not possible since solving for x itself gives me two roots for x (one being $\approx4.79$) and substituting for it in the second equation approx gives me 110. So proving algebraically without any assumptions is not possible. Is this right ?

From your equation you get $5x-x^2=1$, so $5=\frac{x^2+1}{x}$. Now remind the following cubic fromula $(a+b)^3=a^3+b^3+3ab(a+b)$, then
$$
5^3=\left(\frac{x^2+1}{x}\right)^3=\left(x+x^{-1}\right)^3=x^3+x^{-3}+3xx^{-1}(x+x^{-1})=
$$
$$
x^3+x^{-3}+3\frac{x^2+1}{x}=x^3+x^{-3}+15.
$$
So you get
$$
x^3+x^{-3}=125-15=110
$$
