Find (a+b)^4-(a-b)^4. Hence, evaluate (3+2)^4-(3-2)^4 - Vidyadip

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

$( a + b )^4=\left[{ }^4 C_0 a^4+{ }^4 C_1 a^3 b+{ }^4 C_2 a^2 b^2+{ }^4 C_3 a b^3+{ }^4 C_4 b^4\right]$
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]$
$\therefore(a+b)^4-(a-b)^4=2\left[{ }^4 C_1 a^3 b+{ }^4 C_3 a b^3\right]$
$=2\left[4 a^3 b+4 a b^3\right]=8 a b\left[a^2+b^2\right]$
$\therefore(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=8 \cdot \sqrt{3} \cdot \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right]$
$=8 \cdot \sqrt{3} \cdot \sqrt{2}[3+2]$
$=40 \cdot \sqrt{3} \cdot \sqrt{2}$
$=40 \sqrt{6}$

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