Given-d-2-5-1-3-1-5-then-d-3-4d-2-8d-2-

Question Number 194900 by cortano12 last updated on 19/Jul/23

G i v e n d = \frac{\sqrt[3]{2 + \sqrt{5}}}{1 + \sqrt{5}}

t h e n d^{3} - 4 d^{2} + 8 d - 2 = ?

Answered by Frix last updated on 19/Jul/23

d^{3} = \frac{2 + \sqrt{5}}{{(1 + \sqrt{5})}^{3}} = \frac{2 + \sqrt{5}}{8 (2 + \sqrt{5})} = \frac{1}{8} \Leftrightarrow d = \frac{1}{2}

\Rightarrow answer is \frac{9}{8}

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