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

Question

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.

Answer 1

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 $0$.

Answer 2

HINT

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

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

Answer 3

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:

$$x=1$$

Feel free to ask questions about my steps. Hope this helped!