Explanation

(b)
Put $\frac{6 - x + 8 - x}{2} = y \Rightarrow x = 7 - y$
$(y - 1)^{4} + (y + 1)^{4} = 16$
$\Rightarrow y^{4} + 6y^{2} + 1 = 8$
$\Rightarrow y^{4} + 6y^{2} - 7 = 0$
$\Rightarrow y^{2} = 1,\ - 7$
$\Rightarrow y^{2} = 1$ $(\because y^{2} = - 7$ is not possible)
$\Rightarrow y = \pm 1$
$\Rightarrow x = 6,\ 8$
$\therefore$ Total number of real roots are 2.