Answer
The equations hold. See below for proof.
Work Step by Step
(a) We start with the right side:
$\displaystyle \frac{1}{2}[(a+b)^{2}-(a^{2}+b^{2})]=\frac{1}{2}[(a^{2}+b^{2}+2ab)-a^{2}+b^{2}]=\frac{1}{2}(2ab)=ab$
Which equals the left side.
(b) We start with the left side:
$(a^{2}+b^{2})^{2}-(a^{2}-b^{2})^{2}=(a^{2})^{2}+(b^{2})^{2}+2a^{2}b^{2}-[(a^{2})^{2}+(b^{2})^{2}-2a^{2}b^{2}]=4a^{2}b^{2}$
Which equals the right side.
