Why can't we cancel the x from both sides in the equation $x^2=x^3$?

Question

It may seem at first sight that the $x$ can be cancelled from both sides. It doesn't make any sense that the square of a number is equal to its cube (unless it's 1) as in this case. I know that the reason of this is pretty well known. Well, as it happens, my teacher is finding it a bit too difficult in making me understand this.

How can we find the value(s) of $x$ here?Are there complex theorems behind this or is it just simply the 'basics'?

Answer 1

Hint: $\;x^2=x^3 \iff x^3-x^2=0 \iff x^2(x-1)=0\,$, then either factor can be $\,0\,$.

Answer 2

$$x^2-x^3=x^2(1-x)=(x)(x)(1-x)=0$$ so $$x=0$$ $$x=0$$ $$x=1$$

Answer 3

Hint

Isn't $x=0$ a solution too?

Answer 4

$$x^2=x^3$$

First consider whether $x$ can be $0$, $0^2=0^3$. Alright, so $x=0$ is a possible solution.

Next, we consider what if $x \neq 0$, now we can divide by $x^2$.

and the equation become $1=x$.

Hence, in summary, $x=0$ or $x=1$.

Cancelling means dividing here, We have to make sure we do not divide by $0$ accidentally.

Answer 5

We want to find the set $S=\{x\in\mathbb{R}\colon x^2=x^3\}$. Clearly $0\in S$. Now suppose that $x\neq 0$. Then $1/x^2$ exists and $$ x^2=x^3\iff\frac{1}{x^2}x^2=\frac{1}{x^2}x^3\iff x=1. $$ Thus $S=\{0,1\}$.