We can now use Binomial Theorem to evaluate $ (a + b)^4 – (a – b)^4$
$(a+b)^4 = ^4\large C$$_0\;a^nb^0 + ^4\large C$$_1\;a^{4-1}b^1+....$
$(a-b)^4 = ^4\large C$$_0\;a^nb^0 - ^4\large C$$_1\;a^{4-1}b^1+....$
$ (a + b)^4 – (a – b)^4 = ^4\large C$$_0\;a^nb^0 + ^4\large C$$_1\;a^{4-1}b^1+.... ^4\large C$$_4\;a^0b^4 - (^4\large C$$_0\;a^nb^0 - ^4\large C$$_1\;a^{4-1}b^1 + ^4\large C$$_4\;a^{4-4}b^4$)
$\qquad = 2 (\;^4\large C$$_1\;a^3b+\;^4\large C$$_3\;ab^3)$
$\qquad = 2 (4a^3b+4ab^3)$
$\qquad = 8ab(a^2+b^2)$
Therefore, $( \sqrt3 + \sqrt2)^4– (\sqrt 3 –\sqrt 2)^4 = 8 \sqrt 3 \sqrt 2 (\sqrt 3^2 + \sqrt 2^2)$
$\qquad = 8 \times \sqrt 6 \times 5 = 40 \sqrt 6 \approx 97.98$