Find-a-triple-of-rational-numbers-a-b-c-such-that-2-1-3-1-1-3-a-1-3-b-1-3-c-1-3-

Question Number 152281 by john_santu last updated on 27/Aug/21

Find a triple of rational

numbers (a, b, c) such that

\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}

Answered by puissant last updated on 27/Aug/21

t = \sqrt[3]{2} \Rightarrow t^{3} = 2 \to t^{3} - 1 = 1

t - 1 = \frac{t^{3} - 1}{t^{2} + t + 1} = \frac{1}{t^{2} + t + 1} = \frac{3}{3 t^{2} + 3 t + 3}

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

\sqrt[3]{t - 1} = \frac{\sqrt[3]{3}}{t + 1};

t^{3} + 1 = 3,

\sqrt[3]{t - 1} = \frac{\sqrt[3]{3} (t^{2} - t + 1)}{t^{3} + 1} = \frac{\sqrt[3]{3} (t^{2} - t + 1)}{3} = \frac{t^{2} - t + 1}{\sqrt[3]{9}}

t = \sqrt[3]{2} \Rightarrow \sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{4}{9}} + \sqrt[3]{- \frac{2}{9}} + \sqrt[3]{\frac{1}{9}}

∴∵ a = \frac{4}{9}, b = - \frac{2}{9}, c = \frac{1}{9} . .

Commented by Rasheed.Sindhi last updated on 27/Aug/21

N = {1, 2, 3, \dots}

i = \sqrt{- 1}

C = S e t o f C o m p l e x N u m b e r s

e = B a s e o f N a t u r a l L o g a r i t h m

! = S y m b o l f o r f a c t o r i a l