(d) 0
Since $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$, we have $\Sigma f_i x_i = \bar{x} \cdot \Sigma f_i$. Therefore, $\sum f_i(x_i - \bar{x}) = \Sigma f_i x_i - \bar{x}\Sigma f_i = \bar{x}\Sigma f_i - \bar{x}\Sigma f_i = 0$.
Source: Chapter 13, Section 13.2 Mean of Grouped Data
This is a standard property of the arithmetic mean: the sum of deviations of observations from their mean is always zero. Expanding $\sum f_i(x_i - \bar{x}) = \sum f_i x_i - \bar{x}\sum f_i$, and substituting the formula for $\bar{x}$, both terms cancel. Examiners expect you to identify option (d) and optionally show the one-step algebraic justification.