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Question Number 140774 by ajfour last updated on 12/May/21

P r o v e t h a t

{(\frac{1}{\sqrt{2}} + \frac{5}{3 \sqrt{6}})}^{1 / 3} + {(\frac{1}{\sqrt{2}} - \frac{5}{3 \sqrt{6}})}^{1 / 3} = \sqrt{2}

Commented by ajfour last updated on 12/May/21

A l l a r e E x c e l l e n t s o l u t i o n s :

c a n t w e t h e n a l w a y s h a v e a

c o m p a c t a n s w e r t o

{(a + i b)}^{1 / 3} + {(a - i b)}^{1 / 3} = x

w i t h o u t t r i g o f u n c t i o n s, t h a t i s .

Commented by MJS_new last updated on 13/May/21

p^{1 / 3} + q^{1 / 3} = x

p + q + 3 p^{1 / 3} q^{1 / 3} (p^{1 / 3} + q^{1 / 3}) = x^{3}

3 p^{1 / 3} q^{1 / 3} x = x^{3} - p - q

x^{9} - 3 (p + q) x^{6} + 3 (p^{2} - 7 p q + q^{2}) x^{3} - {(p + q)}^{3} = 0

p = a + b i \land q = a - b i

x^{9} - 6 {a x}^{6} - 3 (5 a^{2} + 9 b^{2}) x^{3} - 8 a^{3} = 0

x = t^{1 / 3}

t^{3} - 6 {a t}^{2} - 3 (5 a^{2} + 9 b^{2}) t - 8 a^{3} = 0

t = z + 2 a

z^{3} - 27 (a^{2} + b^{2}) z - 54 a (a^{2} + b^{2}) = 0

only if this has a “ {nice}^{″} solution \dots

Commented by ajfour last updated on 13/May/21

Thanks Sir, i dont think it always be as nice..

Answered by liberty last updated on 12/May/21

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

{(\frac{1}{\sqrt{2}} + \frac{5}{3 \sqrt{6}})}^{1 / 3} + {(\frac{1}{\sqrt{2}} - \frac{5}{3 \sqrt{6}})}^{1 / 3}

= \frac{\sqrt{2}}{{[{(\frac{1}{\sqrt{2}} + \frac{5}{3 \sqrt{6}})}^{2}]}^{1 / 3} - \sqrt[3]{\frac{1}{27}} + {[{(\frac{1}{\sqrt{2}} - \frac{5}{3 \sqrt{6}})}^{2}]}^{1 / 3}}

= \frac{\sqrt{2}}{\sqrt[3]{\frac{52}{54} + \frac{5}{3 \sqrt{3}}} + \sqrt[3]{\frac{52}{54} - \frac{5}{3 \sqrt{3}}} - \frac{1}{3}}

= \frac{\sqrt{2}}{\frac{4}{3} - \frac{1}{3}} = \sqrt{2}

Answered by som(math1967) last updated on 12/May/21

l e t a = (\frac{1}{\sqrt{2}} + \frac{5}{3 \sqrt{6}}) b = (\frac{1}{\sqrt{2}} - \frac{5}{3 \sqrt{6}})

a + b = \frac{2}{\sqrt{2}} = \sqrt{2}

a b = (\frac{1}{2} - \frac{25}{54}) = \frac{2}{54} = \frac{1}{27}

a^{\frac{1}{3}} b^{\frac{1}{3}} = \frac{1}{3}

a + b = \sqrt{2}

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

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

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

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

∴ {(\frac{1}{\sqrt{2}} + \frac{5}{3 \sqrt{6}})}^{1 / 3} + {(\frac{1}{\sqrt{2}} - \frac{5}{3 \sqrt{6}})}^{1 / 3} = \sqrt{2}

Answered by mnjuly1970 last updated on 12/May/21

X^{3} := \sqrt{2} + 3 (\sqrt[3]{\frac{1}{2} - \frac{25}{54}}) (X)

X^{3} = \sqrt{2} + 3 \sqrt[3]{\frac{27 - 25}{54}} X

X^{3} = \sqrt{2} + X \Rightarrow X = \sqrt{2} \in R

X^{3} - 2 X + X - \sqrt{2} = 0

X (X - \sqrt{2}) (X + \sqrt{2}) + (X - \sqrt{2}) = 0

\Rightarrow {_{X^{2} + X \sqrt{2} + 1 = 0 \Rightarrow X \in C}^{X = \sqrt{2}}