Question
Find $(a+b)^4-(a-b)^4$. Hence, evaluate $(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4$
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Answer
$\begin{array}{l}( a + b )^4=\left[{ }^4 C_0 a^4+{ }^4 C_1 a^3 b+{ }^4 C_2 a^2 b^2+{ }^4 C_3 a b^3+{ }^4 C_4 b^4\right] \\
\text { and }(a-b)^4=\left[{ }^4 C_0 a^4-{ }^4 C_1 a^3 b+{ }^4 C_2 a^2 b^2-{ }^4 C_3 a b^3+{ }^4 C_4 b^4\right] \\
\therefore(a+b)^4-(a-b)^4=2\left[{ }^4 C_1 a^3 b+{ }^4 C_3 a b^3\right] \\
=2\left[4 a^3 b+4 a b^3\right]=8 a b\left[a^2+b^2\right] \\
\therefore(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=8 \cdot \sqrt{3} \cdot \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right] \\
=8 \cdot \sqrt{3} \cdot \sqrt{2}[3+2]=40 \cdot \sqrt{3} \cdot \sqrt{2}=40 \sqrt{6}\end{array}$
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