My intuitive side tells me to take the cube root of both the sides and get the answer $1$. However, I realize that it might be a problem for I'll lose solutions as given here:
Is it the case that we always need to have a zero on one side to solve equations like this?
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If you are working in $\mathbb{R}$, then you can easily find that $x = 1$ is a solution, as you have found.
But what about if we wish to find all soutions, including those in $\mathbb{C}$? How do we know that we haven't lost solutions, or that we have all of them?
By the Fundamental Theorem of Algebra, you know that the equation $x^3 = 1$ (or equivalently, $x^3 - 1 = 0$) has three roots. We already know that $x = 1$ is one of them, so you can use your precalculus knowledge of the factor theorem to factor $x - 1$ out to give $(x - 1)(x^2 + x + 1) = 0$. Solving the remaining quadratic gives the solutions $x = \frac{-1 + \sqrt{3}i}{2}$ and $x = \frac{-1 - \sqrt{3}i}{2}$.
You may also be interested in reading up on roots of unity.
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Taking the cubic root of both sides might yield more solutions than you expect. If you consider the equation $x^2=1$, then taking the squareroot yields $\sqrt{x^2}=|x|=\sqrt{1}=1$. Hence $x=\pm 1$. So there are actually two solutions.
Similarly, when taking the cubic root of $x^3=1$ there are three solutions, two of which lie in $\mathbb{C}$.
This is all examsmanship. Is $x$ supposed to be real? Then $1$ is the only possibility. Can $x$ be complex? Then, as shown on the website there are two more solutions.
ADDED: As to how you figure this out, examsmanship includes being able to spot ambiguities in questions -- here "does the exam ask for real roots only or complex ones too?" -- and getting them clarified. If you're told complex, you can think in terms of "this is a cubic equation, it can have two complex solutions and one real, to solve cubics with one known solution you divide by $(x-$known solution) and apply the quadratic formula to the quotient." There are other ways to think about finding the complex roots too. I usually think $\exp(2\pi i/3)$ because I've had exponentials, which you may not have.