# How do I solve for $x$ in the equation $x^3 - 8 = 0$ for which $x \neq 2?$

So my friend gives me a problem.

Solve for $x$ in the following equation: $$x^3 - 8 = 0$$

So I did the following: \begin{align} x^3 \require{cancel}\cancel{- 8} \cancel{+ 8} &= 0 + 8 = 8 \\ x &= \sqrt [3] {8} = 2 \end{align}

Then my friend says, "And...?"

And I say, "What do you mean? I solved for $x$ where $x = 2$."

And my friend says, "And what else does $x$ equal?"

And I say, "$x = \sqrt [3] {8}$ which can only be equal to $2$, so $x$ does not equal anything else...right?"

And my friend says, "No. $x$ does not just equal $2$. What else does $x$ equal?"

And I say, "Uhhh... I don't understand."

I still don't understand. Could you please help me with this? Is this some kind of trick question? If not, what does my friend mean by saying that $x$ does not just equal $2$. Does this mean there is another integer $k$ equal to $\sqrt [3] {8}$ on the condition that $k \neq 2?$ I guess $k$ cannot be an integer then, but it cannot be an imaginary number ($\sqrt{-1}^3 = -\sqrt{-1}$ which is not an integer) so it must be a complex number? How do I approach this problem?

• Question/Hint: Does $x^2 = 9$ imply that $x = \sqrt{9} = 3$? (Or in other words, are there other $x$ values that solve the initial equation?)
– Eff
Oct 8, 2017 at 12:22
• @Eff Yes other values like $-3 = \sqrt{9}$, but that doesn't mean that $-3 = \sqrt [3] {27}$ does it? I am assuming that is what you are implying, if you know what I mean. In other words, are you saying that $-2 = \sqrt [3] {8}?$ Oct 8, 2017 at 12:29
– user223391
Oct 8, 2017 at 12:35
• @user477343 I've given an answer with a general approach to questions of these types.
– Eff
Oct 8, 2017 at 12:40

Clearly, $x=2$ is a solution.
But, $x^3-8=0$ can also be written as,
$x^3-2^3=0$.
This is of the form, $a^3-b^3$.

Proceed further using the formula,
$a^3-b^3=(a-b)(a^2+ab+b^2)$.

You'll get $x=2$ as a solution again, but solve for the quadratic equation using the quadratic formula,
You should be getting two complex solutions.

• This was a very good answer, but I believe Eff's answer was the best. $(+1)$ though :) Oct 9, 2017 at 4:20

As you can check,

$$x^3-8=(x-2)(x^2+2x+4).$$

Then a root is the obvious $x=2$, and the two remaining ones are roots of the quadratic factor. By the well-known formula,

$$x=-1\pm\sqrt{-3}.$$

These roots are not in $\mathbb R$, but in $\mathbb C$. The problem statement should specify if this "counts".

8, like every complex number except for 0, has 3 cube roots. One of them is the principal (real) root, and the other two are complex. The first complex root is approximately $-1 + 1.7321i$, and the other one is approximately $-1 - 1.7321i$.

Hint:

$x^3-8=(x-2)(x^2+2x+4)$. So other two roots (except $x\ne2$ ) are the roots of $(x^2+2x+4)$.

You can solve the latter using quadratic formula.

$x=\frac{-2\pm \sqrt{2^2-4\cdot1\cdot4}}{2\cdot1}$

There is no other real solution to $x^3 - 8 = 0$, other than $x = 2$. You can see this by drawing the graph of $y = x^3-8$, and seeing it only goes through 0 once.

If you want to find the other solutions, you could factorise using a difference of two cubes:

$$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$$ And now you can see where the other solutions are hiding: they must be solutions to $x^2 + 2x + 4 = 0$. We can get these using the quadratic formula, or completing the square:

$$x^2 + 2x + 4 = x^2 + 2x + 1 + 3 = (x + 1)^2 + 3$$

And then, $(x+1)^2 + 3 = 0$ rearranges to $x = 1 \pm \sqrt{-3}$, conventially written as $1 \pm i\sqrt{3}$, where $i$ is some "number" satisfying $i^2 = -1$. You should cubing $1 + i \sqrt{3}$ and seeing that it actually solves the equation.

A classic way to solve such a question is to set $x = re^{\theta i}$ where $r\geq 0$ is the magnitude of $x$, and $\theta$ is its argument. Then $$x^3 = \left(re^{\theta i}\right)^3 = r^3e^{3\theta i} = 8$$

Then $r^3 = |8| = 8$ so $r = 2$, i.e. $x = 2e^{3\theta i}$. Now solve $$e^{3\theta i} = \cos 3\theta + i\sin 3\theta = 1.$$

This implies that $$\cos 3\theta = 1$$ which has solutions $$3\theta = 2\pi k\quad\quad k\in\mathbb Z$$

Hence $\theta = 2\pi k/3$ with $k\in\mathbb Z$. Now find all the different values of $x = r e^{\theta i}$ with $r=2$ and $\theta = 2\pi k/3$ and $k\in\mathbb Z$. Since $e^{i\theta}$ is $2\pi$-periodic there will be $3$ different values.

• By the way, in another answer I've given the general formula for solving the equation $z^n = a$, which is your equation with $a = 8$. See here.
– Eff
Oct 8, 2017 at 12:45
• I think my friend was implying that there was another solution as well, where $x = -1 \pm \sqrt{-3}$ but as we can see, there are more solutions, and I understand well by googling "roots of unity" and looking at the link you provided. You gave all the solutions and not just another solution apart from $2$, so $$(+1) \ \ \ \color{green}{\checkmark}$$ Oct 9, 2017 at 4:18
• What may be “classic” is in the eye of the beholder. I think that the classic way is to divide $X^3-8$ by $X-2$ and look at the quadratic quotient, and see what its roots are. May 22, 2018 at 14:47

It must be specified in which set the roots are allowed.

If $x\in \mathbb Z$ (Diophantine equation) or $x\in \mathbb R$, then there is one root: $x=2$.

If $x\in \mathbb C$, then look for the roots in the form $x=a+bi, a,b\in \mathbb R$:

$$(a+bi)^3=8 \Rightarrow a^3+3a^2bi-3ab^2-b^3i=8+0\cdot i \Rightarrow \\ \begin{cases} a^3-3ab^2=8\\ 3a^2b-b^3=0\end{cases} \Rightarrow \\ 1) \ \begin{cases} a^3-3ab^2=8\\ b=0\end{cases} \qquad \text{or} \qquad \ 2) \ \begin{cases} a^3-3ab^2=8\\ 3a^2-b^2=0\end{cases} \Rightarrow \\ 1) \ (a_1,b_1)=(2,0) \qquad \text{or} \qquad 2) (a_2,b_2)=(-1, -\sqrt{3}); (a_3,b_3)=(-1,\sqrt{3}).$$ Hence, the roots (one real and two complex) are: \begin{align}&x_1=2+0\cdot i=2; \\ &x_2=-1-\sqrt{3}i;\\ &x_2=-1+\sqrt{3}i.\end{align}