If $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$, then the value of $x^6 +\frac{1}{x^6}$ will be :

2702

If $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$,

then the value of $x^6 +\frac{1}{x^6}$ = ?

we know that,

$(x - \frac{1}{x}) = a$

(x² + x^-2) = (a)² + 2 = b

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

According to the question,

$\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}$

then, $x+\frac{1}{x}$ = $\sqrt{6}$² – 2 = 4

and then, $x^3+\frac{1}{x^3}$ = 4³ – 3 × 4 = 52

Now apply the formula of square again,

$x^6 +\frac{1}{x^6}$ = 52² – 2 = 2702