If $x^2-3 x+1=0$, then what is the value of $x^6+\frac{1}{x^6}$ ?

322

x⁴ + x²y² + y⁴ = (x² – xy + y²) (x² + xy + y²)

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

then, $K^2+\frac{1}{K^2}$ = n² – 2

If x + \(\frac{1}{x}\)  = n

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

If $x^2-3 x+1=0$,

then what is the value of $x^6+\frac{1}{x^6}$

Divide If $x^2-3 x+1=0$, by x on both the sides,

x + \(\frac{1}{x}\) = 3

x² + \(\frac{1}{x^2}\) =  3² – 2 = 7

cubing both the sides,

$x^6 +\frac{1}{x^6}$ = 7³ - 3 × 7 = 322