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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'?

1123581321
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crazypanda666
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  • $0^2=0^3=0$ $0×2=0×3$ but $3\ne2$ – hamam_Abdallah Feb 19 '18 at 17:56
  • You can only cancel x's if you are sure $x\neq 0$. – amWhy Feb 19 '18 at 17:56
  • "is it simply the 'basics'": We have a lot of groundwork built up around the theory of polynomials and their roots, most useful here being the fundamental theorem of algebra. By rewording your question of finding the values of $x$ such that $x^2=x^3$ into the question of finding the values of $x$ which are roots to the polynomial equation $f(x)=x^3-x^2$, we can then use all that we know about how real numbers work., i.e. that the roots of $x^3-x^2$ are the roots of $(x-0)^2(x-1)$ which are $0$ and $1$ respectively. – JMoravitz Feb 19 '18 at 18:09

5 Answers5

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

dxiv
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$$x^2=x^3$$

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.

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

E.H.E
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Hint

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

1123581321
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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 $S=\{0,1\}$.

Sri-Amirthan Theivendran
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