<s>[INST] Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
 [/INST] Given equation : $(x^2-3x+1)^2=4x^2-12x+9$  ....(1)
To solve this equation we need the identity,
$a^2-2ab+b^2=(a-b)^2$
on observing the R.H.S. of the equation (1) we can write $4x^2-12x+9$ as
$(2x)^2-2\times2x\times3+3^2$ 
Now from the above identity, it can be written as $(2x-3)^2$
Thus, equation (1) becomes $(x^2-3x+1)^2=(2x-3)^2$
$\implies$ $(x^2-3x+1)^2-(2x-3)^2=0$
$\implies$ $(x^2-3x+1+2x-3)(x^2-3x+1-2x+3)=0$ $\;\;\;\;$$\because\,a^2-b^2=(a+b)(a-b)$
$\implies$ $(x^2-x-2)(x^2-5x+4)=0$
$\implies$ $(x^2-2x+x-2)(x^2-x-4x+4)=0$
$\implies$ $\{x(x-2)+1(x-2)\}\{x(x-1)-4(x-1)\}=0$
$\implies$ $\{(x-2)(x+1)\}\{(x-1)(x-4)\}=0$
$\implies$ $(x-2)(x+1)(x-1)(x-4)=0$
Thus, x = 2, -1, 1, 4 
this is the solution of given equation.
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