How to get the value of $a^{-3} - b^{-3}$

Question

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.

Answer 1

Since $a=1/b$ and $b=1/a$ we get

$$a^{-3}-b^{-3} = b^3-a^3 = (b-a)(b^2+ab+a^2)=-3((a-b)^2+3ab) = -3(9+3)=-36$$

Answer 2

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 $c=-36$

Answer 3

Note that

$$a-b=a-\frac1a=3 \implies \left(a-\frac1a\right)^3=a^3-\frac1{a^3}-3a+\frac3a=\frac1{a^3}-a^3-3\left(a-\frac1a\right)=27\\\implies \frac1{a^3}-a^3=27+3\left(a-\frac1a\right)=36$$

and then

$$a^{-3} - b^{-3}=\frac1{a^3}-a^3=-36$$