If $(x - \frac{1}{x} = 4)$, then what is the value of$(x^6 +\frac{1}{x^6})$ ?
Answer & explanation
Correct answer: option 4
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 $(x - \frac{1}{x} = 4)$
$x^2+\frac{1}{x^2}$ = 42 + 2 = 18
Cubing on both the sides,
$x^6 +\frac{1}{x^6}$ = 183 - 3 × 18 = 5832 - 54
Then the value of$(x^6 +\frac{1}{x^6})$ = 5778