Answer
See prove below.
Work Step by Step
Rewrite the formula in the problem as (take $\mu$ out of the summation as it is a constant)
$\sigma^2=\sum[(X-\mu)^2\cdot P(X)]=\sum[(X^2-2\mu X+u^2)\cdot P(X)]
=\sum X^2P(X)-2\mu\sum XP(X)+\mu^2\sum P(X)$
As $\sum XP(X)=\mu$ and $\sum P(X)=1$,
The above becomes $\sum X^2P(X)-2\mu^2+\mu^2=\sum X^2\cdot P(X)-\mu^2$
which is the same as the formula given on page 267 in this section.
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