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
\((\sqrt3 +\sqrt2)^4 - (\sqrt3 −\sqrt2)^4 = 8(\sqrt3)(\sqrt2)[{(\sqrt3)^2+(\sqrt2)^2]} =8(\sqrt6) [3+2]=40\sqrt6\)

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