y-9-1-3-3-1-3-4-1-1-y-3-

Question Number 170945 by Rasheed.Sindhi last updated on 04/Jun/22

y = \sqrt[3]{9} + \sqrt[3]{3} + 4; {(1 + \frac{1}{y})}^{3} = ?

Answered by cortano1 last updated on 04/Jun/22

y - 3 = \sqrt[3]{9} + \sqrt[3]{3} + 1

(y - 3) (\sqrt[3]{3} - 1) = 2

y = \frac{2}{\sqrt[3]{3} - 1} + 3 = \frac{3 \sqrt[3]{3} - 1}{\sqrt[3]{3} - 1}

{(1 + \frac{1}{y})}^{3} = {(1 + \frac{\sqrt[3]{3} - 1}{3 \sqrt[3]{3} - 1})}^{3}

= {(\frac{4 \sqrt[3]{3} - 2}{3 \sqrt[3]{3} - 1})}^{3} = \frac{48 \sqrt[3]{3} - 96 \sqrt[3]{9} + 184}{9 \sqrt[3]{3} - 27 \sqrt[3]{9} + 80}

Commented by Rasheed.Sindhi last updated on 06/Jun/22

T h a n k y o u s i r!