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Question Number 21877 by FilupS last updated on 06/Oct/17

x =^{3} \sqrt{7 + 5 \sqrt{2}} +^{3} \sqrt{7 - 5 \sqrt{2}}

1 . According to a video, x = 2

2 . According to WolframAlpha,

x \approx 0.2071 + 0.3587 i for “ principal {root}^{″}

and x = 2 \sqrt{2} for “ real - valued {root}^{″}

3 . According to google, x \approx 2.8284

Please help and explain! :)

Commented by Joel577 last updated on 06/Oct/17

{w h e r e}^{'} s t h e v i d e o ?

Commented by Joel577 last updated on 06/Oct/17

Commented by FilupS last updated on 06/Oct/17

Answered by mrW1 last updated on 06/Oct/17

a =^{3} \sqrt{7 + 5 \sqrt{2}}

b =^{3} \sqrt{7 - 5 \sqrt{2}}

x = a + b = real

a^{3} + b^{3} = 14

ab =^{3} \sqrt{(7 + 5 \sqrt{2}) (7 - 5 \sqrt{2})} =^{3} \sqrt{49 - 50} = - 1

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

a^{3} + b^{3} = (a + b) (a^{2} + 2 ab + b^{2} - 3 ab)

a^{3} + b^{3} = (a + b) [{(a + b)}^{2} - 3 ab]

\Rightarrow 14 = x (x^{2} + 3)

\Rightarrow x^{3} + 3 x - 14 = 0

x^{3} - {2 x}^{2} + {2 x}^{2} - 4 x + 7 x - 14 = 0

(x - 2) (x^{2} + 2 x + 7) = 0

\Rightarrow x - 2 = 0

\Rightarrow x = 2

Answered by myintkhaing last updated on 06/Oct/17