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I tried to solve for x in the equation $x^3=x$. I did

$$x^3=x$$

$$x^2=1$$

$$x=\pm1$$

but it's wrong, can anyone help.

N. F. Taussig
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2 Answers2

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You forgot $0$. You shouldn't divide by $x$, or else you will miss it. You can solve it like this:

$$x^3=x$$ subtract $x$ from both sides$$x^3-x=0$$ factor out $x$ $$x(x^2-1)=0$$factor $x^2-1$ $$x(x+1)(x-1)=0$$ Simplify $$x=-1,0,1$$

suomynonA
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The polynomial $p(x) = x^3 - x$ has at most three zeroes by the Fundamental Theorem of Algebra. By inspection, $1, -1, 0$ are three zeroes of $p$: $$ p(1) = 1^3 - 1 = 0 \\ p(-1) = (-1)^3 - (-1) = -1 + 1 = 0 \\ p(0) = 0^3 - 0 = 0, $$ and by the Fundamental Theorem of Algebra, these are the only zeroes of $p$.

Alex Ortiz
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  • Do you want to say a bit more about how you find those zeros? – Thomas Sep 12 '16 at 23:59
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    I don't think it is evident for the OP. – Thomas Sep 13 '16 at 00:40
  • @Thomas Maybe you should consider that the other answer to this question already showed a method of computing the zeroes. The OP may also benefit from seeing someone point out that in certain cases, we don't need to do any hard work if we spend a few extra moments pondering the nature of the problem and other facts we already know. The point of downvoting someone's answer is to send the message that the answer is not useful. In light of others' answers, my answer would be truly useless if I just computed the zeroes like the other answer. – Alex Ortiz Sep 13 '16 at 00:52