(3+√5)^-3 + (3-√5)^-3=?

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(3+√5)^-3 + (3-√5)^-3=?

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Answer 1 · 2018-03-15T14:24:02

$\frac{9}{4}$ $frac(9)(4)$

Explanation:

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

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

Let's combine the fractions:

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

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

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

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

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

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

$= \frac{144}{64}$ $= frac(144)(64)$

$= \frac{9}{4}$ $= frac(9)(4)$

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(3+√5)^-3 + (3-√5)^-3=?

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