if-x-4-1-3-2-1-3-1-find-3-x-3-x-2-1-x-3-

Question Number 150979 by mathdanisur last updated on 17/Aug/21

if x = \sqrt[3]{4} + \sqrt[3]{2} + 1 find \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}} = ?

Answered by john_santu last updated on 18/Aug/21

If x = \sqrt[3]{4} + \sqrt[3]{2} + 1 then \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}} = ?

a^{2} + ab + b^{2} = \frac{a^{3} - b^{3}}{a - b}

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

x = \frac{1}{\sqrt[3]{2} - 1}; \frac{1}{x} = \sqrt[3]{2} - 1

\Leftrightarrow \frac{1}{x} + 1 = \sqrt[3]{2}; {(\frac{1}{x} + 1)}^{3} = 2

\Leftrightarrow \frac{1}{x^{3}} + \frac{3}{x^{2}} + \frac{3}{x} + 1 = 2

\Leftrightarrow \frac{1}{x^{3}} + \frac{3}{x^{2}} + \frac{3}{x} = 1

Commented by bramlexs22 last updated on 17/Aug/21

\frac{3}{x} = 3 \sqrt[3]{2} - 3;

\frac{3}{x^{2}} = 3 {(\sqrt[3]{2} - 1)}^{2} = 3 (\sqrt[3]{4} - 2 \sqrt[3]{2} + 1) = 3 \sqrt[3]{4} - 6 \sqrt[3]{2} + 3

\frac{1}{x^{3}} = {(\sqrt[3]{2} - 1)}^{3} = 1 - 3 \sqrt[3]{4} + 3 \sqrt[3]{2}

∴ \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}} = 3 \sqrt[3]{2} - 3 + 3 \sqrt[3]{4} - 6 \sqrt[3]{2} + 3 + 1 - 3 \sqrt[3]{4} + 3 \sqrt[3]{2}

= - 3 + 3 + 1 = 1

Commented by mathdanisur last updated on 17/Aug/21

Thank you Ser