# Pre-Calculus - Solving for $x^3$

In my Pre-Calculus class we were discussing $i$, and all of it's mysterious imaginary wonders, but I had a moment of confusion during the lecture.

This was the problem: Solve for $x$. $$x^3 - 1 = 0$$

Not too bad; what he said was to factor it: $$(x - 1)(x^2 + x + 1) = 0$$ Then use the quadratic formula for the trinomial, ending up with: $$(x - 1)(\frac{-1 \pm i \sqrt 3} 2)$$ It's pretty straight forward, but then, if you're solving for $x$ in: $$x^2 - 1 = 48$$ You'd move the 1 over and Square Root both sides making $x = \pm 7$. My question is, what are you actually doing if you move the 1 over in the original problem and Cube Root both sides?

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Not really. If you "square root" both sides, you get $|x|=7$, which means $x=7$ or $x=-7$. – Pedro Tamaroff Aug 24 '12 at 19:57
Oh sorry, I meant to put that. – Andre Oseguera Aug 24 '12 at 19:58
When you used the quadratic formula and got the two roots $a=\frac{-1+i\sqrt{3}}{2}$ and $b=\frac{-1-i\sqrt{3}}{2}$ you should be replacing $x^2+x+1$ with $(x-a)(x-b)$, not what you did. – VF1 Aug 24 '12 at 20:07

Not really. If you "square root" both sides, you get $|x|=7$, which means $x=7$ or $x=-7$. Alternatively, you can subtract $48$ to get $$x^2-49=0$$

$$x^2-7^2=0$$

$$(x-7)(x+7)=0$$

$$x-7=0\text{ or } x+7=0$$

The cube root has the "nice" property is is a one one correspondence, from where there is "no harm" if you "cube root" you equation. This means $$x^3=y^3 \iff x=y$$

That is, two real numbers have the same third power if and only if they are the same number in the first place.

The problem with the square root (and in general even powers of $x,y$) is that

$$x^n=y^{n} \iff x=y$$

is false since the additive inverse also works, for example $(-7)^2=7^2$ yet $7=-7$ is manifestly false.

$$x^3-1=0$$

If you're working over $\Bbb R$, that is, considering real numbers as solutions, then there is no harm on doing what you suggest

$$x^3-1=0$$

$$x^3=1\iff x=1$$

Note $-1$ is not a solution since $(-1)^3=-1$. Now, the problem arises when we extend our "workplace" to the complex number system. Then our useful assertion that $$\tag 1 x^3=y^3\iff x=y$$ becomes false. In such case, you must factor the polynomial, as you did, and find complex solutions, since using $(1)$ would be wrong, for it is a true statement for real numbers only. You have found the counterexamples just now $$\left(\frac{-1+i\sqrt 3}{2}\right)^3=\left(\frac{-1-i\sqrt 3}{2}\right)^3=1^3=1$$ but $$\frac{{ - 1 + i\sqrt 3 }}{2} \ne \frac{{ - 1 - i\sqrt 3 }}{2} \ne 1$$

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But if you added the one over and cube root both sides, what would you get? – Andre Oseguera Aug 24 '12 at 20:00
Wouldn't you get $\pm 1$? – Andre Oseguera Aug 24 '12 at 20:02
@AndreOseguera - there two more (complex) solutions as you showed, aside from $+1$, ($-1$ is not a solution) – nbubis Aug 24 '12 at 20:07
$1$ has three (complex) cube roots, just like it has $2$ square roots, and $7$ seventh roots. After a while, you will find out the $n$ $n$-th roots of $1$ are $\cos(2\pi k/n)+i\sin(2\pi k/n)$, $k=0$ to $n-1$. – André Nicolas Aug 24 '12 at 20:08
@AndreOseguera I added something. Let me know. – Pedro Tamaroff Aug 24 '12 at 20:13

All polynomials of degree n has n roots in the complex plane.

In your case the cube root will give you one of three roots. The other two are (as you found out): $$x_2 = \frac{-1+i\sqrt{3}}{2} \\ x_3 = \frac{-1-i\sqrt{3}}{2}$$

All roots to the equation $$x^n = 1, \,\,\, n = 1,2,...$$ are $$e^{\frac{2\pi k i}{n}} = \cos(\frac{2\pi k}{n}) + i\sin(\frac{2\pi k}{n}), \,\, k=0,1,...,n-1$$

In your case $x_1 = 1, x_2=-\frac{1}{2}+i\frac{\sqrt{3}}{2}, x_3=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

These roots are called the n:th roots of unity. http://en.wikipedia.org/wiki/Root_of_unity

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In general, if you solve $x^n=a$ you get $x=\omega \sqrt[n]{a}$ where $\omega$ is one of the n roots of $x^n=1$.

In the case $n=2$ the two possible values $\omega$ takes are $+1$ and $-1$, which lead to the $\pm$ in $\pm \sqrt{a}$.

The general formulas for the n roots of $x^n=1$, called the n-th roots of unity, are

$$\omega_k =\cos \left(\frac{2k \pi}{n}\right)+i \sin \left(\frac{2k \pi}{n}\right) \,;\, 0 \leq k \leq n-1 \,.$$

As soon as you speak of complex numbers in trigonometric form, the above formula becomes obvious and trivial to get.

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My question is, what are you actually doing if you move the 1 over in the original problem and Cube Root both sides?

You are forming an expression $x^3=1$, and then applying the cube root function to both sides. When you do that, you are doing it with the understanding that both sides of the equation are real numbers, because the function you are thinking of is a one-to-one function on the real numbers. So, you only retrieve one answer, "1".

The reason it misses the complex roots is because the whole line of reasoning above relies on the cube function and cuberoot function being inverses of each other on the real numbers, but nothing is said about complex numbers.

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Let us look at it in a different way. The first assumption is that whichever eqn we consider, it is a polynomial equation. When we talk of the solution to the equation we mean the roots of the equation, which means the places where the trajectory of the equation would intersect with the line or plane corresponding to x=0( depending on whether the field of solutions belongs to real or complex numbers.

The eqn associated with $x^2$ can be seen to exist only in real space and so we can neglect the complex space associated with it whereas the polynomial equation involving $x^3$ exists in complex space and for completeness we need to consider the roots in the complex space also.

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