If \(\left(a-\frac 1a\right)^2=4 \) and \(a>1\) what is the absolute value of \(a^3 - \frac1{a^3}?\)

(a - 1/a) = 2 (note that normally we couldn't assume that it isn't -2, but since a > 1, a - 1/a has to be positive.

(a - 1/a)^3 = 8

a^3 - 3a + 3/a - 1/a^3 = 8

a^3 - 1/a^3 - 3(a - 1/a) = 8

Can you figure it out from here?

=^._.^=