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
$$
\sqrt{x^2+x+1}+x^3=\sqrt{2x+2}+x^2+x
$$
 A: A brute force method is to isolated the square roots, $\sqrt{x^2+x+1}-\sqrt{2x+2}=\ldots$ and square both sides. You will still have a square root left; isolate it and square again. This will give you an equation of degree twelve(!), but a few solutions spring to your face: When is and what happens if $\sqrt{2x+2}=0$? When is and what happens if $\sqrt{x^2+x+1}=\sqrt{2x+2}$? WIth the right idea, this brings you down to a degree $7$ polynomial remaining, which - unfortunately - is irreducible.
A: Executing Hagen von Eitzen's program you arrive at the equation
$$(1 + x) (-1 - x + x^2)^2 (1 - x - 5 x^2 - x^3 + 2 x^5 - 3 x^6 + x^7)=0\ .$$
The first two factors produce the zeros
$$x_1=-1, \quad x_2={1\over2}(1-\sqrt{5}),\quad x_3={1\over2}(1+\sqrt{5})\ ,$$
and the third factor has three real zeros that Mathematica computes numerically to
$$x_4=-0.550881,\quad x_5=0.350054,\quad x_6=2.39706\ .$$
But not all of these six $x_k$ are solutions of the original problem. We only have proven that the solution set $S$ is a subset of $\{x_1,x_2,\ldots, x_6\}$.
It is obvious that $x_1\in S$. Furthermore $x^2+x+2=2x+2=3\pm\sqrt{5}$ when $x=x_2$, resp., $x=x_3$. From this observation it then follows that $x_2$, $x_3\in S$ as well. Finally it can be verified numerically that $x_4\in S$, too. The candidates $x_5$ and $x_6$ are definitely not in $S$.
To sum it all up, we have $S=\{x_1,x_2,x_3,x_4\}$.
A: 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.
