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.
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.
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 $$x=-1,0,1$$
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 $p$.