Question

SOLVE IT ...꒯ꄲꋊ'꓄ ꇙꉣꋬꂵ........​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Let

$y = \sqrt{ {x}^{2} - 1 }$

Now

${(x + \sqrt{ {x}^{2} - 1 }) }^{6}$

$= {(x + y)}^{6}$

$\displaystyle \: = {x}^{6} + \binom{6}{1} {x}^{5} y + \binom{6}{2} {x}^{4} {y}^{2} + \binom{6}{3} {x}^{3} {y}^{3} + ...... + \binom{6}{6} {y}^{6}$

Again

${(x - \sqrt{ {x}^{2} - 1 }) }^{6}$

$= {(x - y)}^{6}$

$\displaystyle \: = {x}^{6} - \binom{6}{1} {x}^{5} y + \binom{6}{2} {x}^{4} {y}^{2} - \binom{6}{3} {x}^{3} {y}^{3} + ...... + \binom{6}{6} {y}^{6}$

So

${(x + \sqrt{ {x}^{2} - 1 }) }^{6} + {(x - \sqrt{ {x}^{2} - 1 }) }^{6}$

$= {(x + y)}^{6} + {(x + y)}^{6}$

$\displaystyle \: = 2 \{ {x}^{6} + \binom{6}{2} {x}^{4} {y}^{2} + \binom{6}{4} {x}^{2} {y}^{4}+ \binom{6}{6} {y}^{6} \}$

$\displaystyle \: = 2 ( \: {x}^{6} + 15 \: {x}^{4} {y}^{2} + 15 \: {x}^{2} {y}^{4}+ {y}^{6} )$

$\displaystyle \: = 2 \{ \: {x}^{6} + 15 \: {x}^{4} ( {x}^{2} - 1) + 15 \: {x}^{2} {( {x}^{2} - 1) }^{2} + {( {x}^{2} - 1)}^{3} \}$

$\displaystyle \: = 2 \{ \: {x}^{6} + 15 \: ( {x}^{6} - {x}^{4} ) + 15 \: {x}^{2} ( {x}^{4} - 2 {x}^{2} + 1) + ( {x}^{6} - 3 {x}^{4} + 3 {x}^{2} - 1)\}$

So

The coefficient of

${x}^{4} \: \: \: is \: \:2 \times ( - 15 - 30 - 3) = - 96$

The coefficient of

${x}^{2} \: \: is \: \: 2( 15 + 3) = 36$

RESULT

$\alpha = - 96 \: \: \: and \: \: \beta = 36$

Answer 2

Answer:

When you hear the word condensate, think about condensation and the way gas molecules come together and condense and to a liquid. ... If plasmas are super hot and super excited atoms, the atoms in a Bose-Einstein condensate (BEC) are total opposites. They are super unexcited and super cold atoms.

Step-by-step explanation: