If \({x^6} + \frac{1}{{{x^6}}} = k\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\), then k is equal to

\(\left( {{x^4} + 1 + \frac{1}{{{x^4}}}} \right)\)

\(\left( {{x^2} - 1 + \frac{1}{{{x^2}}}} \right)\)

\(\left( {{x^4} - 1 + \frac{1}{{{x^4}}}} \right)\)

\(\left( {{x^4} - 1 - \frac{1}{{{x^4}}}} \right)\)

If \({x^6} + \frac{1}{{{x^6}}} = k\left( {{x^2} + \frac{1}{{{x^2}}}} \

1.	If \({x^6} + \frac{1}{{{x^6}}} = k\left( {{x^2} + \frac{1}{{{x^2}}}} \right)\), then k is equal to
A.	\(\left( {{x^2} - 1 + \frac{1}{{{x^2}}}} \right)\)
B.	\(\left( {{x^4} - 1 + \frac{1}{{{x^4}}}} \right)\)
C.	\(\left( {{x^4} + 1 + \frac{1}{{{x^4}}}} \right)\)
D.	\(\left( {{x^4} - 1 - \frac{1}{{{x^4}}}} \right)\)
Answer» C. \(\left( {{x^4} + 1 + \frac{1}{{{x^4}}}} \right)\)