Question

Find ${(a+b)}^4+{(a-b)}^4$ evaluate ${(\sqrt{3}+\sqrt{2})}^4+{(\sqrt{3}-\sqrt{2})}^4$ using binomial theorem

Answer 1

$\left({a+b}^n\right){=}_0^{n}c{a}^n{b}^0{+}_1^{n}c{a}^{n-1}{b}^1{+}_2^{n}c{a}^{n-2}{b}^2+.....+.{.}_{n-1}^{n}c{a}^1{b}^{n-1}{+}_n^{n}c{a}^0{b}^n$

$\left({a+b}^4\right){=}_0^{4}c{a}^4{b}^0{+}_1^{4}c{a}^3{b}^1{+}_2^{4}c{a}^2{b}^2{+}_3^{4}c{a}^1{b}^3{+}_4^{4}c{a}^0{b}^4$

$\frac{4}{0(4-0)}{a}^4\times 1+\frac{4}{1\times (4-1)}{a}^2{b}^2+\frac{4}{2(4-2)}{ab}^3+\frac{4}{4(4-4)}1\times b^4$

$\frac{4}{1\times 4}{a}^4+\frac{4}{1\times 3}{a}^{3}b+\frac{4}{2(4-2)}{a}^2{b}^2+\frac{4}{3(4-3)}{ab}^3+\frac{4}{40}{b}^4$

$\left({a+b}^4\right)=a^4+{4a}^{3}b+{6a}^2{b}^2+{4ab}^3{+b}^4$

$\left({a+(-b)}^4\right)=a^4+{4a}^3(-b)+{6a}^2\left({-b}^2\right)+4a\left({-b}^3\right)+\left({-b}^4\right)$

$\left({a+b}^4\right)=a^4-{4a}^{3}b+{6a}^2{b}^2-{4ab}^3+b^4$

$\left({a+b}^4\right)-\left({a-b}^4\right)=a^4-{4a}^{3}b+{6a}^2{b}^2-{4ab}^3+b^4-a^4-{4a}^{3}b+{6a}^2{b}^2-{4ab}^3+b^4$

$\left({a+b}^4\right)-\left({a-b}^4\right)=a^4-{4a}^{3}b+{6a}^2{b}^2-{4ab}^3+b^4-a^4+{4a}^{3}b-{6a}^2{b}^2+{4ab}^3-b^4$

$\left({a+b}^4\right)-\left({a-b}^4\right)=a^4-a^4+{6a}^2{b}^2-{6a}^2{b}^2+b^4-b^4{+4ab}^3{+4ab}^3+{4a}^{3}b+{4ab}^3$

$0+0+0+{8a}^{3}b{+8ab}^3$

${8a}^{3}b{+8ab}^3$

$8ab(a^2+b^2)$

$\left({a+b}^4\right)-\left({a-b}^4\right)=8ab(a^2+b^2)$

${(\sqrt{3}+\sqrt{2})}^4-{(\sqrt{3}-\sqrt{2})}^4,a=\sqrt{3},b=\sqrt{2}$

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

$8\sqrt{3}\sqrt{2}$

$8\sqrt{3}\sqrt{2}\left(5\right)$

$8\times 5\sqrt{3\times 2}$

$40\times \sqrt{6}$