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Expansion of (x^2 - y^2)

√x+1+√x-16+√x...

Question

$(\sqrt{x + 1} + \sqrt{x - 1})^{6} + (\sqrt{x + 1} - \sqrt{x - 1})^{6}$ .

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Solution

${(\sqrt{x + 1} + \sqrt{x - 1})}^{6} + {(\sqrt{x + 1} - \sqrt{x - 1})}^{6}$
${({(\sqrt{x + 1} + \sqrt{x - 1})}^{2})}^{3} + {({(\sqrt{x + 1} - \sqrt{x - 1})}^{2})}^{3}$
$= [{(\sqrt{x + 1} + \sqrt{x - 1})}^{2} + {(\sqrt{x + 1} - \sqrt{x - 1})}^{2}]$
$({(\sqrt{x + 1} + \sqrt{x - 1})}^{4} + {(\sqrt{x + 1} - \sqrt{x - 1})}^{4} - (x + 1 - (x - 1)))$
$= (x + 1 + x - 1 + 2 \sqrt{x^{2} - 1} + x + 1 + x - 1 - 2 \sqrt{x^{2} - 1})$
$[{({[\sqrt{x + 1} + \sqrt{x - 1}]}^{2})}^{2} + {({[\sqrt{x + 1} - \sqrt{x - 1}]}^{2})}^{2} - 2]$
$= 4 x [{(x + 1 + x - 1 + 2 \sqrt{x^{2} - 1})}^{2} + {(x + 1 + x - 1 - 2 \sqrt{x^{2} - 1})}^{2} - 2]$
$= 4 x [{(2 x + 2 \sqrt{x^{2} + 1})}^{2} + {(2 x - 2 \sqrt{x^{2} - 1})}^{2} - 2]$
$= 4 x [4 [x^{2} + x^{2} + 1 + 2 x \sqrt{x^{2} + 1} + x^{2} + x^{2} - 1 - 2 x \sqrt{x^{2} - 1}] - 2]$
$= 4 x (4 (4 x^{2}) - 2)$
$= 4 x (16 x^{2} - 2)$
$= 64 x^{3}$

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x + \frac{1}{x} = 2

x^{16} + \frac{1}{x^{16}}

\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1

{(1 + x + x^{2})}^{8} = a_{0} + a_{1} x + . . . . . a_{16} x^{16}

a_{0} - a_{2} + a_{4} - a_{6} + . . . . . + a_{16} =

x

\frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, - 1

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