if-x-7-6-1-3-7-6-1-3-find-E-x-x-6-6x-4-9x-2-

Question Number 151317 by mathdanisur last updated on 20/Aug/21

if x = \sqrt[3]{\sqrt{7} - \sqrt{6}} + \sqrt[3]{\sqrt{7} + \sqrt{6}}

find E (x) = x^{6} - {6 x}^{4} + {9 x}^{2} = ?

Answered by iloveisrael last updated on 20/Aug/21

w e h a v e x - \sqrt[3]{\sqrt{7} - \sqrt{6}} - \sqrt[3]{\sqrt{7} + \sqrt{6}} = 0

i m p l i e s t h a t x^{3} - (\sqrt{7} - \sqrt{6}) - (\sqrt{7} + \sqrt{6}) = 3 x \sqrt[3]{7 - 6}

\Rightarrow x^{3} - 2 \sqrt{7} = 3 x

\Rightarrow x^{3} - 3 x = 2 \sqrt{7}

\Rightarrow {(x^{3} - 3 x)}^{2} = 28

\Rightarrow x^{6} - 6 x^{4} + 9 x^{2} = 28

Answered by som(math1967) last updated on 20/Aug/21

l e t a =^{3} \sqrt{\sqrt{7} - \sqrt{6}}, b =^{3} \sqrt{\sqrt{7} + \sqrt{6}}

∴ a b =^{3} \sqrt{(\sqrt{7} - \sqrt{6}) (\sqrt{7} + \sqrt{6})}

=^{3} \sqrt{7 - 6} = 1

a^{3} + b^{3} = \sqrt{7} - \sqrt{6} + \sqrt{7} + \sqrt{6} = 2 \sqrt{7}

n o w x = a + b

x^{3} = {(a + b)}^{3}

x^{3} = a^{3} + b^{3} + 3 a b (a + b)

x^{3} = 2 \sqrt{7} + 3.1 . x [x = a + b]

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

{(x^{3} - 3 x)}^{2} = {(2 \sqrt{7})}^{2}

x^{6} - 6 x^{4} + 9 x^{2} = 28 a n s

Commented by mathdanisur last updated on 20/Aug/21

Thank You S er