Find the sum of squares of the real solutions to $\sqrt[3]{5x - 2} = \frac{1}{5}(x^3 + 2)$

Question

Find the sum of squares of the real solutions to $\sqrt[3]{5x - 2} = \frac{1}{5}(x^3 + 2)$

I tried applying Viete's formulas after raising everything to the third power, however computing $\sum_{k = 1}^9 x_k^2$ also includes complex solutions, yet the problem only asks us for real solutions.

One can easily spot that $x_1 = 2$ is a solution, and Wolfram Alpha states the other two solutions are $x_2 = \sqrt{2} - 1$ and $x_3 = -\sqrt{2} - 1$, therefore the sum should be 10. However, how could I prove this mathematically?

Thank you very much for your help!

Answer 1

$\sqrt[{\mathrm{3}}]{\mathrm{5}{x}−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}}\left({x}^{\mathrm{3}} +\mathrm{2}\right) \\ $ $\Leftrightarrow\mathrm{5}\sqrt[{\mathrm{3}}]{\mathrm{5}{x}−\mathrm{2}}={x}^{\mathrm{3}} +\mathrm{2} \\ $ $\mathrm{Let}:\:{a}=\sqrt[{\mathrm{3}}]{\mathrm{5}{x}−\mathrm{2}}\Rightarrow{a}^{\mathrm{3}} =\mathrm{5}{x}−\mathrm{2} \\ $ $\mathrm{From}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{have} \\ $ $\begin{cases}{\mathrm{5}{a}={x}^{\mathrm{3}} +\mathrm{2}}\\{\mathrm{5}{x}={a}^{\mathrm{3}} +\mathrm{2}}\end{cases} \\ $ $\Rightarrow\mathrm{5}\left({a}−{x}\right)={x}^{\mathrm{3}} −{a}^{\mathrm{3}} =\left({x}−{a}\right)\left({x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} \right) \\ $ $\Leftrightarrow\left({a}−{x}\right)\left({x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} +\mathrm{5}\right)=\mathrm{0} \\ $ $\Leftrightarrow{a}={x}\:\mathrm{or}\:{x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} +\mathrm{5}=\mathrm{0} \\ $ $\mathrm{But}:\:{x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} +\mathrm{5}=\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{5}\neq\mathrm{0} \\ $ $\Rightarrow{a}={x}\Leftrightarrow{x}^{\mathrm{3}} =\mathrm{5}{x}−\mathrm{2}\Leftrightarrow{x}^{\mathrm{3}} −\mathrm{5}{x}+\mathrm{2}=\mathrm{0} \\ $ $\Leftrightarrow{x}=\mathrm{2},{x}=−\mathrm{1}\pm\sqrt{\mathrm{2}}\Rightarrow\mathrm{Sum}\:\mathrm{of}\:\mathrm{squares}=\mathrm{10} \\ $

Answer 2

Here $\displaystyle f(x)=\frac{x^3+2}{5}.$ Then $\displaystyle f^{-1}(x)=(5x-2)^{\frac{1}{3}}$

So we have to find number of solution of $\displaystyle f(x)=f^{-1}(x)$

Which equals $f(x)=x\Longrightarrow x^3+2=5x$

(Because inverse function is symmetrical about $y=x$ line)

So $\displaystyle x^3-5x+2=0\Longrightarrow x=2\,\ ,-1\pm \sqrt{2}$