Question

Find $(a + b)^4 - (a - b)^4$. Hence, evaluate $(\sqrt3 + \sqrt2)^4 - (\sqrt3 - \sqrt2)^4$.

Answer 1

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

$(a+b)^4 = ^4C_0 a^4+ ^4C_1a^3b+ ^4C_2a^2b^2 + ^4C_3ab^3 + ^4C_4b^4$
$(a-b)^4 = ^4C_0 a^4- ^4C_1a^3b+ ^4C_2a^2b^2 - ^4C_3ab^3 + ^4C_4b^4$
∴$(a+b)^4-(a - b)^4 = ^4C_0 a^4+ ^4C_1a^3b+ ^4C_2a^2b^2 + ^4C_3ab^3 + ^4C_4b^4$ $-$ $[^4C_0 a^4- 4C_1a^3b+ ^4C_2a^2b^2 - ^4C_3ab^3 + ^4C_4b^4\,]$
=$2(^4C_1 a^3b+ ^4C_3ab^3)$
=$2(4a^3b+4ab^3)$
=$8ab (a^2+b^2)$

By putting $a = \sqrt3$ and $b = \sqrt2$, we obtain
$(\sqrt3 +\sqrt2)^4 - (\sqrt3 −\sqrt2)^4 = 8(\sqrt3)(\sqrt2)[{(\sqrt3)^2+(\sqrt2)^2]} =8(\sqrt6) [3+2]=40\sqrt6$

$\text{Hence,} \,(\sqrt3 + \sqrt2)^4 - (\sqrt3 - \sqrt2)^4=40\sqrt6.$