Question

question of class 11, binomial theorem​

Answer 1

$\orange{\mathbb{BINOMIAL THEOREM}} \\ \\ \tt We\:know\:that \\ \\ \tt \boxed{\bold{\underline{\green{\tt {(x+y)}^{n} = {}^{n}{C}_{0}{x}^{n}{y}^{0} + {}^{n}{C}_{1}{x}^{n-1}{y}^{1} + ... + {}^{n}{C}_{n}{x}^{0}{y}^{n} }}}} \\ \\ \tt \implies So\:{(a+b)}^{4} = {}^{4}{C}_{0}{a}^{4}{b}^{0} + {}^{4}{C}_{1}{a}^{4-1}{y}^{1} + {}^{4}{C}_{2}{a}^{4-2}{b}^{2} + {}^{4}{C}_{3}{a}^{4-3}{b}^{3} + {}^{4}{C}_{4}{a}^{4-4}{b}^{4} \\ \\ \tt \implies {(a+b)}^{4} = \frac{4!}{4!0!}{a}^{4} + \frac{4!}{(4-1)!1!}{a}^{3}b + \frac{4!}{(4-2)!2!} {a}^{2}{b}^{2} + \frac{4!}{(4-3)!3!}a{b}^{3} + \frac{4!}{0!4!}{b}^{4} \\ \\ \tt \therefore \boxed{\bold{\underline{\purple{\tt {(a+b)}^{4} = {a}^{4} + 4{a}^{3}b + 6{a}^{2}{b}^{2} + 4a{b}^{3} + {b}^{4} }}}} \\ \\ \tt \implies So\:{(a-b)}^{4} = {}^{4}{C}_{0}{a}^{4}{-b}^{0} + {}^{4}{C}_{1}{a}^{4-1}{-b}^{1} + {}^{4}{C}_{2}{a}^{4-2}{-b}^{2} + {}^{4}{C}_{3}{a}^{4-3}{-b}^{3} + {}^{4}{C}_{4}{a}^{4-4}{-b}^{4} \\ \\ \tt \implies {(a-b)}^{4} = \frac{4!}{4!0!}{a}^{4} - \frac{4!}{(4-1)!1!}{a}^{3}b + \frac{4!}{(4-2)!2!} {a}^{2}{b}^{2} - \frac{4!}{(4-3)!3!}a{b}^{3} + \frac{4!}{0!4!}{b}^{4} \\ \\ \tt \therefore \boxed{\bold{\underline{\purple{\tt {(a-b)}^{4} = {a}^{4} - 4{a}^{3}b + 6{a}^{2}{b}^{2} - 4a{b}^{3} + {b}^{4} }}}} \\ \\ \tt \implies {(a+b)}^{4} - {(a-b)}^{4} \\ \\ \tt \implies( {a}^{4} + 4{a}^{3}b + 6{a}^{2}{b}^{2} + 4a{b}^{3} + {b}^{4} )-({a}^{4} - 4{a}^{3}b + 6{a}^{2}{b}^{2} - 4a{b}^{3} + {b}^{4}) \\ \\ \tt \implies {a}^{4} + 4{a}^{3}b + 6{a}^{2}{b}^{2} + 4a{b}^{3} + {b}^{4} - {a}^{4} + 4{a}^{3}b - 6{a}^{2}{b}^{2} + 4a{b}^{3} - {b}^{4} \\ \\ \tt \implies 8{a}^{3}b + 8a{b}^{3}\\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt {(a+b)}^{4} - {(a-b)}^{4} = 8{a}^{3}b + 8a{b}^{3} }}}} \\ \\ \tt {(\sqrt{3}+\sqrt{2})}^{4} - {(\sqrt{3}-\sqrt{2})}^{4} \\ \\ \tt \implies 8{(\sqrt{3})}^{3}\sqrt{2} + 8\sqrt{3}{(\sqrt{2})}^{3} \\ \\ \tt \implies 8*3\sqrt{6} + 8*2\sqrt{6} \\ \\ \tt \implies 24\sqrt{6} + 16\sqrt{6} \\ \\ \tt \implies 40\sqrt{6} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt {(\sqrt{3}+\sqrt{2})}^{4} - {(\sqrt{3}-\sqrt{2})}^{4} = 40\sqrt{6} }}}}$