Solve for x :
$x^4-6x^3+12x^2-12x+4=0$

Tried to factorize and used substitution but no result.

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There's no equation there, so there's nothing to solve. – user61527 Oct 17 '13 at 6:43

I’m assuming that the problem is to solve

$$x^4-6x^3+12x^2-12x+4=0\;.$$

Here’s a very elementary approach that will work if there is a nice solution. Since there is no rational root, we have no easy way to find a root $r$ and divide out $x-r$ to reduce the problem to that of solving a cubic equation. We can hope, though, that the lefthand side is a product of the form

$$(x^2+ax+c)(x^2+bx+d)\;,\tag{1}$$

where clearly $cd=4$. If we’re really fortunate, the coefficients will be integers. For no good reason I decided to try first the possible factorization

$$(x^2+ax+2)(x^2+bx+2)\;,$$

setting it equal to the original quartic and equating coefficients:

$$(x^2+ax+2)(x^2+bx+2)=x^4+(a+b)x^3+(ab+4)x^2+2(a+b)x+4\;,$$

so

\left\{\begin{align*} &a+b=-6\\ &ab+4=12\\ &2(a+b)=-12\;. \end{align*}\right.

The third equation is equivalent to the first, so we want to solve the system

\left\{\begin{align*} &a+b=-6\\ &ab=8\;. \end{align*}\right.

It has a nice solution, so I’ll leave the rest to you, since it’s all quite straightforward. Note that I was fortunate: I picked the right factorization right off the bat. Had I tried

$$(x^2+ax+1)(x^2+bx+4)\;,$$

for instance, I’d have found myself with an inconsistent system. (Of course I was extremely lucky that there actually is a factorization of the form $(1)$ with integer $c$ and $d$.)

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