If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for

  • A

    All real $x$

  • B

    ${x^2} \le 2$

  • C

    ${x^2} \ge 2$

  • D

    None of these

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