Question

Find $(a+b{)}^4-(a-b{)}^4$. Hence find $s$ if $(\sqrt{3}+\sqrt{2}{)}^4-(\sqrt{3}-\sqrt{2}{)}^4=40\sqrt{s}$

Answer 1

Using Binomial theorem, the expressions $(a+b{)}^4$ and $(a-b{)}^4$ can be expanded as,
$(a+b{)}^4{=}^4{C}_0{a}^4{+}^4{C}_1{a}^{3}b{+}^4{C}_2{a}^2{b}^2{+}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4)$
$(a-b{)}^4{=}^4{C}_0{a}^4{-}^4{C}_1{a}^{3}b{-}^4{C}_2{a}^2{b}^2{-}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4)$
$\therefore (a+b{)}^4-(a-b{)}^4$ =${}^4{C}_0{a}^4{+}^4{C}_1{a}^{3}b{+}^4{C}_2{a}^2{b}^2{+}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4-{[}^4{C}_0{a}^4{-}^4{C}_1{a}^{3}b{-}^4{C}_2{a}^2{b}^2{-}^4{C}_{3}a{b}^3{+}^4{C}_4{b}^4]$
$=$ $2{(}^4{C}_1{a}^{3}b{+}^4{C}_{3}a{b}^3)=2(4a^{3}b+4a{b}^3)$
$=$ $8ab(a^2+b^2)$
Now by putting $a=\sqrt{3}$ and $b=\sqrt{2}$, we obtain
$(\sqrt{3}+\sqrt{2}{)}^4-(\sqrt{3}-\sqrt{2}{)}^4=8\left(\sqrt{3}\right)\left(\sqrt{2}\right)\left\{\right(\sqrt{3}{)}^2+\left(\sqrt{2}{)}^2\right\}=8\left(\sqrt{6}\right)\{3+2\}=40\sqrt{6}$