# (3+√5)^-3 + (3-√5)^-3=?

Mar 15, 2018

$\frac{9}{4}$

#### Explanation:

We have: ${\left(3 + \sqrt{5}\right)}^{- 3} + {\left(3 - \sqrt{5}\right)}^{- 3}$

$= \frac{1}{{\left(3 + \sqrt{5}\right)}^{3}} + \frac{1}{{\left(3 - \sqrt{5}\right)}^{3}}$

Let's combine the fractions:

$= \frac{{\left(3 - \sqrt{5}\right)}^{3} + {\left(3 + \sqrt{5}\right)}^{3}}{{\left(3 + \sqrt{5}\right)}^{3} {\left(3 - \sqrt{5}\right)}^{3}}$

$= \frac{\left({3}^{3} + 3 \cdot {\left(3\right)}^{2} \cdot - \sqrt{5} + 3 \cdot 3 \cdot {\left(- \sqrt{5}\right)}^{2} + {\left(- \sqrt{5}\right)}^{3}\right) + \left({3}^{3} + 3 \cdot {\left(3\right)}^{2} \cdot \sqrt{5} + 3 \cdot 3 \cdot {\left(\sqrt{5}\right)}^{2} + {\left(\sqrt{5}\right)}^{3}\right)}{{\left(\left(3 + \sqrt{5}\right) \left(3 - \sqrt{5}\right)\right)}^{3}}$

We can apply the algebraic identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ to the denominator:

$= \frac{\left(27 - 27 \sqrt{5} + 45 - 5 \sqrt{5}\right) + \left(27 + 27 \sqrt{5} + 45 + 5 \sqrt{5}\right)}{{\left({3}^{2} - {\left(\sqrt{5}\right)}^{2}\right)}^{3}}$

$= \frac{54 + 90}{{\left(9 - 5\right)}^{3}}$

$= \frac{144}{{4}^{3}}$

$= \frac{144}{64}$

$= \frac{9}{4}$