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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:

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Is it the case that we always need to have a zero on one side to solve equations like this?

Irregular User
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MathEnthusiast
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  • See http://math.stackexchange.com/questions/192742/how-to-solve-x3-1 – lab bhattacharjee Apr 21 '16 at 07:00
  • If you haven't studied complex numbers then the answer by mathematicsstudent is all you need. Yet the link you post shows all the solutions in $;\Bbb C;$ , so: have you studied complex numbers? – DonAntonio Apr 21 '16 at 07:04
  • Yes I have. But I never thought of subtracting both sides so that I can factor it later. My intuition told me to take the cube root. How was I supposed to figure out that I was to subtract on both the sides instead of doing it the former way? – MathEnthusiast Apr 21 '16 at 07:06
  • @user331377 One of the first things that are usually taught when studying complex numbers in high school (at least here) is to calculate roots of unity of degree $;n;$ ,which means solutions to $;x^n=1;$, using the polar form of complex numbers, de Moivre's and Euler's formula and etc. – DonAntonio Apr 21 '16 at 07:11
  • Yeah, but it's not given in this book. – MathEnthusiast Apr 21 '16 at 07:12

3 Answers3

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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.

Irregular User
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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.

ForgotALot
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  • So, how will I know that I am supposed to subtract both the sides by one and then factorize instead of just taking the cube root on both the sides? – MathEnthusiast Apr 21 '16 at 07:05
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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 $\mathbb{C}$.

Mathematician 42
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  • I tend to take both paths; for $x^3=1$, I take the cube root of both sides to identify a solution, and then divide (maybe with long division) by $(x - \text{one solution})$ to simplify the expression. My goal here is to factor the expression completely. – John Joy Apr 21 '16 at 22:39