When to neglect the negative value of a square root?

Question

The question I was trying to solve is:

$$If\ X^{4}+(\frac{1}{X})^{4}=119, then\ \ prove\ \ that\ \ X^{3}-(\frac{1}{X})^{3}=36$$

I got the following solution from internet:

$$x^4+\frac{1}{x^4}=119 $$

$$ \left(x^2+\frac{1}{x^2}\right)^2= \displaystyle x^4+\frac{1}{x^4}+2$$

$$\implies$$

$$ \left(x^2+\frac{1}{x^2}\right)^2= \displaystyle 119+2=121$$

$$ \left(x^2+\frac{1}{x^2}\right)^2= \displaystyle (11)^2$$

$$ \boxed{\left(x^2+\frac{1}{x^2}\right)= 11} \tag 1$$

$$ \left(x-\frac{1}{x}\right)^2= \displaystyle x^2+\frac{1}{x^2}-2$$

$$\implies$$

$$ \left(x-\frac{1}{x}\right)^2= 11-2=9$$

$$ \left(x-\frac{1}{x}\right)^2= (3)^2$$

$$\boxed{\displaystyle \left(x-\frac{1}{x}\right) = 3} \tag 2$$ $$\left(x-\frac{1}{x}\right)^3 = (3)^3$$ $$x^{3}-\frac{1}{x^3}-3x^{3}.\frac{1}{x^3}(x-\frac{1}{x})=27$$ $$x^{3}-\frac{1}{x^3}-3(3)=27$$ $$x^{3}-\frac{1}{x^3}-9=27$$ $$x^{3}-\frac{1}{x^3}=27+9$$ $$x^{3}-\frac{1}{x^3}=36$$

Please notice $(1)$ and $(2)$. The solution is neglecting the negative value of square root which are $-11$ and $-3$. I'm trying to understand WHY?

Answer 1

In the first case, it's because $x^2+\frac1{x^2}>0$. In the second case, that should not have been done. Actually, the statement$$X^4+\left(\frac1X\right)^4=119\implies X^3-\left(\frac1X\right)^3=36$$is false. Even assuming that $X\in\Bbb R$, all that you can deduce is that$$X^3-\left(\frac1X\right)^3=\pm36.$$

Answer 2

For (1), it's due to the non-negativeness of the square of reals. For (2), I can see no reason either.