Explanation

(b)
Now, we assume $(b - c)^{2},\ (c - a)^{2},\ (a - b)^{2}$ are in AP, then we have
$(c - a)^{2} - (b - c)^{2} = (a - b)^{2} - (c - a)^{2}$
$\Rightarrow (b - a)(2c - a - b) = (c - b)(2a - b - c)$ …(i)
Also, if $\frac{1}{b - c},\frac{1}{c - a},\frac{1}{a - b}$ are in AP, then
$\frac{1}{c - a} - \frac{1}{b - c} = \frac{1}{a - b} - \frac{1}{c - a}$
$\Rightarrow \ \ \frac{b + a - 2c}{(c - a)(b - c)} = \frac{c + b - 2a}{(a - b)(c - a)}$
$\Rightarrow (a - b)(b + a - 2c) = (b - c)(c + b - 2a)$
$\Rightarrow \ \ (b - a)(2c - a - b) = (c - b)(2a - b - c)$
Which is equal to Eq. (i), so, our hypothesis is true.