If $x^{2} + 6x+ 1 = 0$, then the value of $\left(x + 6\right)^{3} + \frac{1}{\left(x + 6\right)^{3}}$ = ?

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If x + \(\frac{1}{x}\)  = n

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

If $x^{2} + 6x+ 1 = 0$

Find $\left(x + 6\right)^{3} + \frac{1}{\left(x + 6\right)^{3}}$

Let x + 6 = t 

x = t - 6

Put the value of x in $x^{2} + 6x+ 1 = 0$

$(t-6)^{2} + 6(t-2)+ 1 = 0$

= t² + 36 - 12t + 6t - 12 + 1 = 0

= t² - 6t + 25 = 0

= t(t-6) + 25 = 0

= t ×