Solve for real solution(s) for [MATH]x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)[/MATH].

Solve for real solution(s) for [MATH]x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)[/MATH].

My solution:

Let [MATH]y=(x+1)^2=x^2+2x+1[/MATH], this turn the given equality into the following form:

[MATH]x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)[/MATH]

[MATH]\underline{x^2+2x+ 1} -3x= (\underline{x^2+2x+ 1}-x)(\underline{x^2+2x+ 1}+ 3)[/MATH]

[MATH]y -3x= (y-x)(y+ 3)[/MATH]

Expanding we get:

[MATH]y -3x= y^2+3y-xy-3x[/MATH]

[MATH]y = y^2+3y-xy[/MATH]

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

[MATH](-1)^2− (-1) + 1 = ((-1)^2+ (-1) + 1)((-1)^2+ 2(-1) + 4)\implies 3=3[/MATH]

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

[MATH]1 = y+3-x[/MATH]

[MATH]x-2= y[/MATH]

[MATH]x-2=x^2+2x+1[/MATH]

[MATH]x^2+x+3=0[/MATH]

[MATH]\left(x-\frac{1}{2}\right)^2+3-\frac{1}{4}=0[/MATH]

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

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