if-a-4-1-3-2-1-3-1-find-3-a-3-a-2-1-a-3-

Question Number 160190 by HongKing last updated on 25/Nov/21

if a = \sqrt[3]{4} + \sqrt[3]{2} + 1

find \frac{3}{a} + \frac{3}{a^{2}} + \frac{1}{a^{3}} = ?

Answered by amin96 last updated on 25/Nov/21

a = \frac{{(\sqrt[3]{2})}^{3} - 1}{\sqrt[3]{2} - 1} = \frac{1}{\sqrt[3]{2} - 1}

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

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

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

Answered by mr W last updated on 25/Nov/21

a = \sqrt[3]{4} + \sqrt[3]{2} + 1

a + 1 = \sqrt[3]{4} + \sqrt[3]{2} + 2

a + 1 = \sqrt[3]{4} + \sqrt[3]{2} + \sqrt[3]{4} \sqrt[3]{2}

a + 1 = (\sqrt[3]{2} + 1 + \sqrt[3]{4}) \sqrt[3]{2}

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

{(1 + \frac{1}{a})}^{3} = 2

1 + \frac{3}{a} + \frac{3}{a^{2}} + \frac{1}{a^{3}} = 2

\Rightarrow \frac{3}{a} + \frac{3}{a^{2}} + \frac{1}{a^{3}} = 1

Answered by 1549442205PVT last updated on 16/Dec/21

A p p l y a^{3} - b^{3} = (a - b) {(a^{2} + a b + b)}^{2} w e g e t

a = \sqrt[3]{4} + \sqrt[3]{2} + 1 \Rightarrow \frac{1}{a} = \frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1}

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

= \frac{\sqrt[3]{2} - 1}{{(^{3} \sqrt{2})}^{3} - 1} =^{3} \sqrt{2} - 1 . H e n c e, \frac{1}{a} + 1 =^{3} \sqrt{2}

\frac{3}{a} + \frac{3}{a^{2}} + \frac{1}{a^{3}} = {(\frac{1}{a} + 1)}^{3} - 1 = {(^{3} \sqrt{2})}^{3} - 1 = 1