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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

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1 Answer

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

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