# Computing $\sqrt[3]{1\,}$

I know that the answer is always $1$, but they are looking for some way to get to that answer and I don't know what it is. I am not good at english math terms, but maybe it has to do with differential functions or something like that.

• In a more-general setting, $\sqrt[3]{1}$ is actually multi-valued (in other words, it isn't necessarily always $1$). – Clayton Aug 27 '15 at 19:56
• Compute $(-{1 \over 2} + i {\sqrt{3} \over 2})^3$. More generally, consider $1=e^{i 2 \pi}$. – copper.hat Aug 27 '15 at 19:57
• Are you working with $\mathbb{R}$ or $\mathbb{C}$? – copper.hat Aug 27 '15 at 20:02
• I am working with R – lakistrike Aug 27 '15 at 20:07
• Henry's answer should suffice. Also note that if $x^3=1$, then $|x|^3 = 1$ and so $|x| = 1$. There are only two possibilities on the real line. – copper.hat Aug 27 '15 at 20:10

You could try to solve $x^3=1$, i.e. $x^3-1=0$ which you can factorise to $(x-1)(x^2+x+1)=0$ which implies $x-1=0$ or $x^2+x+1=0$.

$x-1=0$ gives $x=1$, as expected, while $x^2+x+1 = (x+\frac12)^2 + \frac34$ is strictly positive for any real $x$ and so not $0$.

• Another example is ∛(〖27.9〗^2 ) – lakistrike Aug 27 '15 at 20:14
• and then this formula is used: f(x+Δx)≈ f(x)+f '(x)*Δx – lakistrike Aug 27 '15 at 20:17
• My mistake, it is √(3.92^3) – lakistrike Aug 27 '15 at 20:25

## HINT

Let $x = \sqrt[3]{1}$. You'll have:

$$x^3 = 1 \\ x^3 - 1 = 0 \\ (x - 1)(x^2 + x + 1) = 0$$

Sometimes roots can trip people up. Try this:

$$\sqrt[3]{1}=x$$

$$1^{\frac{1}{3}}=x$$

$$1^{\frac{1}{3}*{\frac{3}{1}}}=x^{\frac{3}{1}}$$

$$1=x^3$$

Now, you should know that $1$ times itself is always $1$. So if I multiply $1$ by itself $3$ times, or in other words raise $1$ to the power of $3$, I'll still have $1$, right? Plug in $1$ for $x$ to check this:

$$1=(1)^3$$

$$1=1$$

Yep, it works. Therefore:

$$x=1$$