Solve the equation $$\sqrt{x^{2}+ 8}- \sqrt{x^{2}+ 3}+ 2x^{3}- x- 2= 0$$
My solution is $$x+ 2- 2x^{3}= \sqrt{x^{2}+ 8}- \sqrt{x^{2}+ 3}= \frac{5}{\sqrt{x^{2}+ 8}+ \sqrt{x^{2}+ 3}}\leq \frac{5}{\sqrt{8}+ \sqrt{3}}< \frac{3}{2}$$ $$\Rightarrow 2x^{3}- x- \frac{1}{2}> 0\Rightarrow \frac{(8x- 7)\left ( 64x^{2}+ 56x+ 17 \right )}{256}> \frac{9}{256}> 0\Rightarrow x> \frac{7}{8}$$ We always have that $$\sqrt{x^{2}+ 8}- \sqrt{x^{2}+ 3}+ 2x^{3}- x- 2= 0$$ $$\Leftrightarrow \left ( \sqrt{x^{2}+ 8}- 3 \right )- \left ( \sqrt{x^{2}+ 3}- x- 1 \right )\left ( x^{3}+ 2x^{2}+ x+ 1+ x(x+ 1)\sqrt{x^{2}+ 3} \right )= 0$$ $$\Leftrightarrow (x- 1)\left ( \frac{x+ 1}{\sqrt{x^{2}+ 8}+ 3}+ \frac{x^{3}+ 2x^{2}+ x+ 1+ x(x+ 1)\sqrt{x^{2}+ 3}}{\sqrt{x^{2}+ 3}+ x+ 1} \right )= 0$$ The answer is $x= 1$. I'll leave the rest for you. I need to a fresh solution for the above one. Thanks..
Another solution due to my friend
We always have that $$\sqrt{x^{2}+ 8}- \sqrt{x^{2}+ 3}+ 2x^{3}- x- 2= 0$$ $$\begin{matrix} \Rightarrow \frac{1}{x+ 1}\left ( \sqrt{x^{2}+ 8}- \sqrt{x^{2}+ 3}- 1 \right )\\ \left ( 2x^{4}+ 2x^{3}+ 11x^{2}+ 8x+ 3+ (2x^{2}+ 2x+ 1)\left ( 2\sqrt{x^{2}+ 8}+ 3\sqrt{x^{2}+ 3}+ \sqrt{x^{2}+ 8}\sqrt{x^{2}+ 3} \right ) \right )= 0 \end{matrix}$$ On the other hand $$2x^{4}+ 2x^{3}+ 11x^{2}+ 8x+ 3= 2\left ( x^{2}+ \frac{x}{2} \right )^{2}+ \frac{21}{2}\left ( x+ \frac{8}{21} \right )^{2}+ \frac{31}{21}> 0$$ Solved