Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$

My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4-7x^3-8x^2-1=0 $$
 A: Set $A=x^2,B=x^2+x+1$. 
Then,
$$\begin{align}10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2&=10A^2-7AB+B^2\\&=(2A-B)(5A-B)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}$$
A: 
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$

Set $t=x^2,z=x^2+x+1$.
$\Longrightarrow$
$$\begin{align}10t^2-7tz+z^2&=(2t-z)(5t-z)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}$$
$$\boxed{\color{red}{x_{1,2}=\frac{1}{2}\pm\frac{\sqrt5}{2},\;x_{3,4}=\frac{1}{8}\pm \frac{\sqrt{17}}{8}}}$$
A: Divide by $(x^2 + x + 1)^2$, the equation becomes: $$10\frac{x^4}{(x^2 + x + 1)^2} - 7\frac{x^2}{x^2 + x + 1} + 1 = 0$$ Let $z = \frac{x^2}{x^2 + x + 1}$. The equation now is $$10z^2 - 7z + 1 = 0$$ Solving it like an ordinary quadratic equation on $z$ you get at most two roots $z_{1,2}$. Then let $\frac{x^2}{x^2 + x + 1} = z_1$, or, equivalently, $$x^2 = z_1 (x^2+x+1)$$ It is a quadratic equation in $x$. Similary for $z_2$. Solutions to this two equations are solutions to the initial problem.
A: $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$
Divide by $x^4$ on both sides,
$10-\frac{7x^2(x^2+x+1)}{x^4}+\frac{(x^2+x+1)^2}{x^4}=0$
$10-7(1+\frac{1}{x}+\frac{1}{x^2})+(1+\frac{1}{x}+\frac{1}{x^2})^2=0$
Put $(1+\frac{1}{x}+\frac{1}{x^2})=t$
$10-7t+t^2=0$,solving we get $t=2,5$
when $t=2$
$1+\frac{1}{x}+\frac{1}{x^2}=2$
simplify we get,$x^2-x-1=0$.............(1)
when $t=5$
$1+\frac{1}{x}+\frac{1}{x^2}=5$
simplify we get,$4x^2-x-1=0...........(2)$
Solve (1) and (2) and get the answer.
