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

Let's solve the given equation:

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

First, we identify that we can factor the expression on the left side using factoring out the common factor:

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

We'll continue and factor the expression in parentheses, we can identify that it can be factored using the perfect square trinomial formula:

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

Let's do it:

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

We should emphasize that this factoring using the mentioned formula was possible only because the middle term in the expression (which is in first power in this case and highlighted in blue in the previous calculation) indeed matched the middle term in the perfect square trinomial formula,

We'll continue and get two simpler equations and solve them:

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

et's summarize the equation solving steps:

$$$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.