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Solve the equation $$(x+3)(\sqrt{2x^2+6x+2}-2x)=\sqrt[3]{x^2+1}+(x^2-7)\sqrt{x+3}$$

It's little complicated, any help will be appreciated.

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  • $\begingroup$ The solve command of Maple produces $-\sqrt{7},\,\sqrt{7}$. $\endgroup$ – user64494 Aug 12 '15 at 20:26
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$$ (x+3)\sqrt{2x^2+6x+2}-(2x^2+6x+2)=\sqrt[3]{x^2+1}-2+(x^2-7)\sqrt{x+3}$$

$$\Rightarrow \sqrt{2x^2+6x+2}\frac{7-x^2}{x+3+\sqrt{2x^2+6x+2}}=\frac{x^2-7}{\sqrt[3]{(x^2+1)^2}+2\sqrt[3]{x^2+1}+4}+(x^2-7)\sqrt{x+3}$$

And so the solutions $${x\in\{-\sqrt 7,+\sqrt 7\}}$$

And $$-\frac{\sqrt{2x^2+6x+2}}{x+3+\sqrt{2x^2+6x+2}}=\frac{1}{\sqrt[3]{(x^2+1)^2}+2\sqrt[3]{x^2+1}+4}+\sqrt{x+3}$$ Which clearly has no more solution since $LHS\le 0<RHS$

AND WE'RE DONE

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