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

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

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

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

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

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

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 $$x=2\pm\sqrt{2}$$.

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