Solve for real solution(s) for $\displaystyle x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$.

Solve for real solution(s) for $\displaystyle x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$.

My solution:

Let $\displaystyle y=(x+1)^2=x^2+2x+1$, this turn the given equality into the following form:

$\displaystyle x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$

$\displaystyle \underline{x^2+2x+ 1} -3x= (\underline{x^2+2x+ 1}-x)(\underline{x^2+2x+ 1}+ 3)$

$\displaystyle y -3x= (y-x)(y+ 3)$

Expanding we get:

$\displaystyle y -3x= y^2+3y-xy-3x$

$\displaystyle y = y^2+3y-xy$

If $y=0$, then $x=-1$ is a solution, and we verified that it indeed is a solution:

$\displaystyle (-1)^2− (-1) + 1 = ((-1)^2+ (-1) + 1)((-1)^2+ 2(-1) + 4)\implies 3=3$

If $y\ne 0$, then divide through the equation $\displaystyle y = y^2+3y-xy$ by $y$ we get:

$\displaystyle 1 = y+3-x$

$\displaystyle x-2= y$

$\displaystyle x-2=x^2+2x+1$

$\displaystyle x^2+x+3=0$

$\displaystyle \left(x-\frac{1}{2}\right)^2+3-\frac{1}{4}=0$

$\displaystyle \left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\implies$ no real solution from this case.

Therefore, the only real solution for the given equation is $x=-1$.