Solve for x: $(x^2-3x+1)^2-3(x^2-3x+1)+1=x$

Question

Solve the equation for $x$: $$(x^2-3x+1)^2-3(x^2-3x+1)+1=x$$

I've tried to solve by:

Let $y=(x^2-3x+1)$

So, $$y^2-3y+1=x$$

But still have one $x$ and it's not factorable!

What do I do to solve this?

Thank you

Answer 1

Hint: you are halfway there.

$$ x^2 - 3x + 1 = y \\ y^2 - 3y + 1 = x $$

Subtract:

$$(y-x)(y+x-2)=0$$

The latter gives two quadratics in $x$.

Answer 2

Hint: $y- x= x^2 -y^2-3(x-y)$ . Can you continue?