Explanation

(b)
Since, $\sin{\theta + \cos{\theta = x}}$ …(i)
and $\sin^{6}{\theta + \cos^{6}{\theta = \frac{1}{4}\lbrack 4 - 3\left( x^{2} - 1 \right)^{2}\rbrack}}$
On equation Eq (i), we get
$\sin{2\theta = x^{2} - 1 \leq 1\ \ \ \ (\because\sin{2\theta \leq 1)}}$
$\Rightarrow x^{2} \leq 2 \Rightarrow - \sqrt{2} \leq x \leq \sqrt{2}$
Now, $\sin^{6}{\theta + {\cos^{6}{\theta\ = (\sin^{2}{\theta + \cos^{2}{\theta)}}}}^{3} - 3\sin^{2}\theta}\cos^{2}{\theta(sin^{2}\theta + \cos^{2}{\theta)}}$
$= 1 - 3\sin^{2}{\theta\cos^{2}{\theta = 1 - \frac{3}{4}\sin^{2}{2\theta}}}$
$= 1 - \frac{3}{4}\left( x^{2} - \right)^{2} = \frac{1}{4}\lbrack 4 - 3\left( x^{2} - 1 \right)^{2}\rbrack$
Thus, the given result will hold true only when $x^{2} \leq 2$ and not for all real values of $x$