<s>[INST] How to get the value of $a^{-3} - b^{-3}$ I have a statement that says:

If $a - b = 3$, and $a$ is the reciprocal of $b$, that is equal to $a = \frac{1}{b}$, get the value of $a^{-3} - b^{-3}$

So i have,
$a = 3 + b$, $b = a - 3$, $a = \frac{1}{b}$
$3 + b  = \frac{1}{b}$, then ($b^2 + 3b - 1 = 0$ )
or 
$a = \frac{1}{a-3}$, then ($a^2 -3a -1 = 0$ )
So, i need the value of $a^{-3} - b^{-3}$
Now i use the Quadratic formula for $a$:
$a = \frac{3 \pm\sqrt{13}}{2}$, then $a^-3 = \frac{8}{(3\pm\sqrt{13})^3}$
Now i use the Quadratic formula for $b$:
$b = \frac{-3 \pm\sqrt{13}}{2}$, then $b^-3 = \frac{8}{(-3\pm\sqrt{13})^3}$
Finally,
= $\frac{8}{(3\pm\sqrt{13})^3} - \frac{8}{(-3\pm\sqrt{13})^3}$, I will use $+$ instead of the $\pm$, so
= $\frac{8}{(3+\sqrt{13})^3} - \frac{8}{(-3+\sqrt{13})^3}$
= $\frac{8}{40\sqrt{13}+144} - \frac{8}{40\sqrt{13}-144}$
= $-36$
In fact, the result is correct but the development, has been very long, for a question that I am supposed to answer in less than 2 minutes, So, what other way do you see to solve this exercise? I've been looking for it on my own, but I do not clarify with another form.
 [/INST] So $a=b^{-1}$ and therefore $c=a^{-3}-b^{-3}=a^{-3}-a^{3}$
Now $a-b=a-\frac 1a$ and $(a-\frac 1a)^3=a^3-a^{-3}-3(a-\frac 1a)$ so that $27=-c-9$ and $c=-36$
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