Solve the equation $\sqrt[3]{15-x^3+3x^2-3x}=2\sqrt{x^2-4x+2}+3-x$.

I have tried to solve for x by Casio and try to make the equation to $u.v=0$ but the solution is not in $\mathbb{Q}$. Any help is appreciated. Thanks

  • 1
    $\begingroup$ potentially interesting rewrite, but now sure how useful yet $$\sqrt[3]{14 - (x-1)^3} = 2\sqrt{(x-2)^2-2} - (x-3)$$ $\endgroup$ – gt6989b Feb 13 at 16:23
  • $\begingroup$ I have rewrite this but I'm stuck now $\endgroup$ – Nguyen Thy Feb 13 at 16:31

For the square root to be defined we need:

$$x^2-4x+2\geq 0$$

Therefore, we have:

$$2\sqrt{x^2-4x+2} = x-3+\sqrt[3]{15-x^3+3x^2-3x}=$$


$$=\frac{-6(x^2-4x+2)}{(x-3)^2-(x-3)\sqrt[3]{15-x^3+3x^2-3x}+\sqrt[3]{(15-x^3+3x^2-3x)^2}}\leq 0$$

The numerator is negative $x^2-4x+2 \geq 0 \Rightarrow -6(x^2-4x+2)\leq 0$. The denominator is of the form $a^2-ab+b^2$ which is always non-negative because:

$$a^2-ab+b^2=\frac{1}{2}[a^2+b^2+(a-b)^2]\geq 0$$

And thus $x^2-4x+2=0$ which means $x=2\pm\sqrt{2}$.

  • $\begingroup$ Can you explain more why it's $\le 0$? $\endgroup$ – Nguyen Thy Feb 13 at 16:47
  • $\begingroup$ @NguyenThy, Because the numerator is negative. $x^2-4x+2 \geq 0 \Rightarrow -6(x^2-4x+2)\leq 0$ $\endgroup$ – LHF Feb 13 at 16:49
  • $\begingroup$ How about determinator ? $\endgroup$ – Nguyen Thy Feb 13 at 16:50
  • $\begingroup$ @NguyenThy, I edited to add more detail. $\endgroup$ – LHF Feb 13 at 16:51
  • $\begingroup$ I see. Thank you very much. $\endgroup$ – Nguyen Thy Feb 13 at 16:53

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