Question
Obtain the value of $(\sqrt{3}+2)^{5}-(\sqrt{3}-2)^{5}$ using binomial expansion method.

Answer

$ (\sqrt{3}+2)^{5}-(\sqrt{3}-2)^{5} $
$ =\left[\begin{array}{c} { }^{5} C_{0}(\sqrt{3})^{5}(2)^{0} \\ +{ }^{5} C_{1}(\sqrt{3})^{4}(2)^{1} \\ +{ }^{5} C_{2}(\sqrt{3})^{3}(2)^{2} \\ +{ }^{5} C_{3}(\sqrt{3})^{2}(2)^{3} \\ +{ }^{5} C_{4}(\sqrt{3})^{1}(2)^{4} \\ +{ }^{5} C_{5}(\sqrt{3})^{0}(2)^{5} \end{array}\right]-\left[\begin{array}{c} { }^{5} C_{0}(\sqrt{3})^{5}(2)^{0} \\ -{ }^{5} C_{1}(\sqrt{3})^{4}(2)^{1} \\ +{ }^{5} C_{2}(\sqrt{3})^{3}(2)^{2} \\ -{ }^{5} C_{3}(\sqrt{3})^{2}(2)^{3} \\ +{ }^{5} C_{4}(\sqrt{3})^{1}(2)^{4} \\ -{ }^{5} C_{5}(\sqrt{3})^{0}(2)^{5} \end{array}\right] $
(As there is a negative sign between two binomial expressions, the signs of terms of the second expression will change. As a result, the first, third and the fifth terms in the expansion of both expressions will get cancelled.)
$ \begin{aligned} &=2\left[{ }^{5} \mathrm{C}_{1}(\sqrt{3})^{4}(2)^{1}+{ }^{5} \mathrm{C}_{3}(\sqrt{3})^{2}(2)^{3}+{ }^{5} \mathrm{C}_{5}(\sqrt{3})^{0}(2)^{5}\right] \\ &=2[5(9)(2)+10(3)(8)+32] \\ &=2[90+240+32] \\ &=2[362] \\ &=724 \end{aligned} $

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