Answer
$a)$ $ab=\frac{1}{2}[(a+b)^{2}-(a^{2}+b^{2})]$
$b)$ $(a^{2}+b^{2})^{2}-(a^{2}-b^{2})^{2}=4a^{2}b^{2}$
Work Step by Step
$a)$ $ab=\frac{1}{2}[(a+b)^{2}-(a^{2}+b^{2})]$
Evaluate $(a+b)^{2}$ inside the brackets:
$ab=\dfrac{1}{2}\Big[a^{2}+2ab+b^{2}-(a^{2}+b^{2})\Big]$
Simplify the expression inside the brackets:
$ab=\dfrac{1}{2}\Big[a^{2}+2ab+b^{2}-a^{2}-b^{2}\Big]$
$ab=\dfrac{1}{2}[2ab]$
Evaluate the product on the right side and the identity will be proved:
$ab=ab$
$b)$ $(a^{2}+b^{2})^{2}-(a^{2}-b^{2})^{2}=4a^{2}b^{2}$
Evaluate both powers on the left:
$(a^{4}+2a^{2}b^{2}+b^{4})-(a^{4}-2a^{2}b^{2}+b^{4})=4a^{2}b^{2}$
Simplify the left side and the identity will be proved:
$a^{4}+2a^{2}b^{2}+b^{4}-a^{4}+2a^{2}b^{2}-b^{4}=4a^{2}b^{2}$
$2a^{2}b^{2}+2a^{2}b^{2}=4a^{2}b^{2}$
$4a^{2}b^{2}=4a^{2}b^{2}$
