How to find roots of $f(x)=x^3-2$

Question

How do I find the roots of $x^3-2=0$? I have used the rational root theorem but it just isn't working, and all the methods I have used have yielded me no proper result. What do I do?

Answer 1

The rational root theorem does not work because there are no rational roots.

When all fails, use graphing, from which you can determine the real roots of any equation (as an approximation). In this case, $x\approx 1.26$, which is a good estimate.

Here is a graph, where the parent function is $x^3$, so this function is just translated 2 units down:

Answer 2

$x^3-b^3=(x-b)(x^2+bx+b^2)$, hence $x^3-2=(x-\sqrt[3]{2})(x^2+ \sqrt[3]{2}x + 2^{\frac{2}{3}})$.

The discriminant of the quadratic is negative and hence has no real roots. However, the linear factor has a real root. Hence, the real roots of $x^3 - 2=0$ is $x=\sqrt[3]{2}$

Answer 3

$$ x^3 - 2 = 0 $$ $$ x^3 = 2 $$ $$ x = 2^{\frac{1}{3}} $$