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Question Number 43793 by gunawan last updated on 15/Sep/18

The value of^{3} \sqrt{20 + 14 \sqrt{2}} +^{3} \sqrt{20 - 14 \sqrt{2}} is \dots

Answered by Joel578 last updated on 15/Sep/18

Let a = 20 + 14 \sqrt{2}

b = 20 - 14 \sqrt{2}

a b = (20 + 14 \sqrt{2}) (20 - 14 \sqrt{2}) = 400 - 392 = 8

\sqrt[3]{a} + \sqrt[3]{b} = x

a + 3 \sqrt[3]{a^{2} b} + 3 \sqrt[3]{{a b}^{2}} + b = x^{3}

(a + b) + 3 \sqrt[3]{a b} (\sqrt[3]{a} + \sqrt[3]{b}) = x^{3}

(40) + 3 \sqrt[3]{8} (x) = x^{3}

40 + 6 x = x^{3}

x^{3} - 6 x - 40 = 0

(x - 4) (x^{2} + 4 x + 10) = 0

∴ x = 4

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18

20 + 14 \sqrt{2}

= 2^{3} + 3 . 2^{2} . (\sqrt{2}) + 3.2 . {(\sqrt{2})}^{2} + {(\sqrt{2})}^{3}

= {(2 + \sqrt{2})}^{3}

20 - 14 \sqrt{2} = {(2 - \sqrt{2})}^{3}

{(20 + 14 \sqrt{2})}^{\frac{1}{3}} + {(20 - 14 \sqrt{2})}^{\frac{1}{3}}

= 2 + \sqrt{2} + 2 - \sqrt{2}

= 4