Solve x³-2x²+x=0: Finding All Solutions of a Cubic Equation

Let's solve the given equation:

$x^3-2x^2+x=0$

Note that we can factor the expression on the left side by factoring out the common factor:

$$$x^3-2x^2+x=0 \\ \downarrow\\ x(x^2-2x+1)=0$

Proceed to factor the expression inside of the parentheses. It can be factored by using the perfect square trinomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$

As shown below:

$x(x^2-2x+1)=0 \\ x(x^2 \textcolor{blue}{-2x}+1^2)=0 \\ x(x^2\textcolor{blue}{-2\cdot x \cdot 1}+1^2)=0 \\ \downarrow\\ x(x-1)^2=0$

We should emphasize that the process of factoring by using the mentioned formula was only possible due to the middle term in the expression. (The first power in this case is highlighted in blue indeed matched the middle term in the perfect square trinomial formula)

Having obtained two simpler equations let's proceed to solve them:

$x(x-1)^2=0 \\ \boxed{x=0}\\ (x-1)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\rightarrow x-1=0\rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Shown below is a summary of the various steps to solve the given equation:

$$$x^3-2x^2+x=0 \\ \downarrow\\ x(x^2-2x+1)=0 \\ \downarrow\\ x(x-1)^2=0 \\ \downarrow\\ \boxed{x=0}\\ (x-1)^2=0\rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore, the given equation has two different solutions,

Which means - the correct answer is answer B.