Question

Find $(a+b{)}^4-(a-b{)}^4$. Hence, evaluate $(\sqrt{3}+\sqrt{2}{)}^4-(\sqrt{3}-\sqrt{2}{)}^4$.

Answer 1

$(a+b{)}^4-(a-b{)}^4$

$=[4_{C_0}{a}^4+4_{C_1}{a}^{3}b+4_{C_2}{a}^2{b}^2+4_{C_3}a{b}^3+4_{C_4}{b}^4]$

$-[4_{C_0}{a}^4+4_{C_1}{a}^3(-b)+4_{C_2}{a}^2(-b{)}^2+4_{C_3}a(-b{)}^3+4_{C_4}(-b{)}^4]$

$=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4-[a^4-4a^{3}b+6a^2{b}^2-4a{b}^3+b^4]$

$=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4-a^4+4a^{3}b-6a^2{b}^2+4a{b}^3-b^4$

Putting $a=\sqrt{3}$ and $b=\sqrt{2}$

$(\sqrt{3}+\sqrt{2}{)}^4-(\sqrt{3}-\sqrt{2}{)}^4=8\sqrt{3}.\sqrt{2}\left[\right(\sqrt{3}{)}^2+\left(\sqrt{2}{)}^2\right]$

$=8.\sqrt{6}[3+2]=40\sqrt{6}$