Find (a + b)4 – (a – b)4. Hence, evaluate.

Using Binomial Theorem, the expressions, $(a+b)^{4}$ and $(a-b)^{4}$, can be expanded as

$(a+b)^{4}={ }^{4} C_{0} a^{4}+{ }^{4} C_{1} a^{3} b+{ }^{4} C_{2} a^{2} b^{2}+{ }^{4} C_{3} a b^{3}+{ }^{4} C_{4} b^{4}$

$(a-b)^{4}={ }^{4} C_{0} a^{4}-{ }^{4} C_{1} a^{3} b+{ }^{4} C_{2} a^{2} b^{2}-{ }^{4} C_{3} a b^{3}+{ }^{4} C_{4} b^{4}$

$\therefore(a+b)^{4}-(a-b)^{4}={ }^{4} C_{1} a^{4}+{ }^{4} C_{1} a^{3} b+{ }^{4} C_{2} a^{2} b^{2}+{ }^{4} C_{3} a b^{3}+{ }^{4} C_{1} b^{4}$

$-\left[{ }^{4} \mathrm{C}_{0} \mathrm{a}^{4}-{ }^{4} \mathrm{C}_{1} \mathrm{a}^{3} \mathrm{~b}+{ }^{4} \mathrm{C}_{2} \mathrm{a}^{2} \mathrm{~b}^{2}-{ }^{4} \mathrm{C}_{3} \mathrm{ab}^{3}+{ }^{4} \mathrm{C}_{4} \mathrm{~b}^{4}\right]$

$=2\left({ }^{4} \mathrm{C}_{1} \mathrm{a}^{3} \mathrm{~b}+{ }^{4} \mathrm{C}_{3} \mathrm{ab}^{3}\right)=2\left(4 \mathrm{a}^{3} \mathrm{~b}+4 \mathrm{ab}^{3}\right)$

$=8 \mathrm{ab}\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)$

By putting $a=\sqrt{3}$ and $b=\sqrt{2}$, we obtain

$(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}=8(\sqrt{3})(\sqrt{2})\left\{(\sqrt{3})^{2}+(\sqrt{2})^{2}\right\}$

$=8(\sqrt{6})\{3+2\}=40 \sqrt{6}$