How to solve this simple problem So my younger brother asked me to solve a simple problem, but I'm not sure how to solve this. Any help would be much appreciated. 
$$a^3 - 1/ a^3 = 4. $$
Prove that  $$a - 1/a = 1$$
 A: Using  $$\displaystyle \left(a-\frac{1}{a}\right)^3=\left(a^3-\frac{1}{a^3}\right)-3\cdot a\cdot \frac{1}{a}\left(a-\frac{1}{a}\right)$$
So we get $$\displaystyle \left(a-\frac{1}{a}\right)^3=4-3\displaystyle \left(a-\frac{1}{a}\right)\;,$$ Now Put $\displaystyle \left(a-\frac{1}{a}\right)=x$
So we get $$\displaystyle x^3=4-3x\Rightarrow x^3+3x-4=0\Rightarrow (x-1)\cdot (x^2+x+4)=0$$
So we get $x=1$ or $\displaystyle x^2+x+4 =0\Rightarrow x=\frac{-1\pm \sqrt{1-16}}{2}$ (No real values of $x$)
A: By reverse approach solve quadratic equation 
$$ x - \frac{1}{x} =4 \rightarrow  x_{1,2} = 2 \pm \sqrt 5 $$
checks ok because
$$ \frac {1}{x_1} = \frac{1}{ 2 + \sqrt 5} = - 2 + \sqrt 5 $$
and plugging into cubic relation, I get after simplifying,
$$ x_1^3 -\frac{1}{x_1^3}= 76 ,\,  x_2^3 -\frac{1}{x_2^3}= 76 \,  !! $$
A: So let's solve the first problem for $a$ in the reals.
$a^3-a^{-3}=4$. If thrown into mathematica you get the real solutions are $\frac{1}{2}(1\pm \sqrt{5})$. Then solving the second for $a$ in the reals gives us $a-a^{-1}=1$. This also has the same solutions. Alternatively if we were to assume $a>0$ both simplify to $a^2=a+1$.
