Prove-that-2-5-1-3-2-5-1-3-is-a-rational-number-

Question Number 18386 by Tinkutara last updated on 19/Jul/17

Prove that \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} is a

rational number .

Commented by mrW1 last updated on 19/Jul/17

\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1

Answered by mrW1 last updated on 20/Jul/17

x = \sqrt[3]{2 + \sqrt{5}}

y = \sqrt[3]{2 - \sqrt{5}}

let u = x + y

xy = \sqrt[3]{(2 + \sqrt{5}) (2 - \sqrt{5})} =^{3} \sqrt{4 - 5} =^{3} \sqrt{- 1} = - 1

x^{3} + y^{3} = (2 + \sqrt{5}) + (2 - \sqrt{5}) = 4

\Rightarrow (x + y) (x^{2} + y^{2} - xy) = 4

\Rightarrow (x + y) (x^{2} + y^{2} + 2 xy - 3 xy) = 4

\Rightarrow (x + y) [{(x + y)}^{2} - 3 xy] = 4

\Rightarrow u (u^{2} + 3) = 4

u^{3} + 3 u - 4 = 0

u^{3} - u + 4 u - 4 = 0

(u^{2} - 1) u + 4 (u - 1) = 0

(u - 1) [u (u + 1) + 4] = 0

(u - 1) (u^{2} + u + 4) = 0

(u - 1) [{(u + \frac{1}{2})}^{2} + \frac{15}{4}] = 0

\Rightarrow u = 1

i . e . \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1

Commented by Tinkutara last updated on 20/Jul/17

Thanks Sir!