<s>[INST] Any one help me solve Algebraically equation $\sqrt{x^2+x+1}+ x^{3}= \sqrt{2x+2}+x^{2}+x$ I can't solve this. Can any one show me detail solution
$$
\sqrt{x^2+x+1}+x^3=\sqrt{2x+2}+x^2+x
$$
 [/INST] This can be solved easily by observing the aid equation once you rearrange the terms. 
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} = -x^3 + x^2 + x
$$
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} = -x(x^2 -x -1)
$$
Notice that on the LHS, the difference between squares of the terms in under the root is RHS/$x$
Let's multiply and divide by the conjugate on the LHS
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} \times \frac{(\sqrt{x^2 + x + 1} + \sqrt{2x+2})}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} = -x(x^2 -x -1)
$$
$$
\frac{x^2 -x -1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}}= -x(x^2 -x -1)
$$
Rearranging a few terms,
$$
\frac{x^2 -x -1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x(x^2 -x -1) = 0
$$
$$
(x^2 -x -1) \times \left ( \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x \right ) = 0
$$
Now, either of $ (x^2 -x -1) , \Bigl ( \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x \Bigr)$ could be trivial.
We know that $x \geq -1$ (Constraint placed on $ \sqrt{2x + 2}$ ) making $ \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x  = 0$ at $ x = -1$
The other solutions lie in $ x^2 -x -1 = 0$, which can be solved as a standard quadratic polynomial. 
These 3 values will be your solution.
Edit : There is a another solution inside the equation, which can be found by solving $ \Bigl ( \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x \Bigr) = 0$ completely.
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