Question

Find the value of $(2-\sqrt{1-x}{)}^6+(2+\sqrt{1-x}{)}^6$.

Answer 1

${\left(2-\sqrt{1-x}\right)}^6+{\left(2+\sqrt{1-x}\right)}^{6}let\sqrt{1-x}=a{\left(2-a\right)}^6+{\left(2+a\right)}^{6}Byusingbionomialression\left({}^6{C}_0{a}^0{.2}^6{-}^6{C}_1{a}^1{.2}^5{+}^6{C}_2{a}^2{.2}^4-{}^6{C}_3{a}^3{.2}^3{+}^6{C}_4{a}^4{.2}^2{-}^6{C}_5{a}^5{.2}^1+{}^6{C}_6.a^6{.2}^0\right)+\left({}^6{C}_o{a}^0{.2}^6{+}^6{C}_1{a}^1{.2}^5{+}^6{C}_2{a}^2{.2}^4{+}^6{C}_{3}a{.}^3{2}^3+{}^6{C}_4{a}^4{.2}^2{+}^6{C}_5{a}^5{.2}^1{+}^6{C}_6{a}^6{.2}^0\right)2\left({}^6{C}_o{.2}^6{+}^6{C}_2{a}^2{.2}^4{+}^6{C}_4.a^4{.2}^2{+}^6{C}_6{a}^6{.2}^0\right)2\left(64+\frac{6\times 5}{2\times 1}{a}^2\times 16+\frac{6\times 5}{2\times 1}{a}^4.4+1\times a^6\right)\left[\because a=\sqrt{1-x}\right]2{\left[64+240a^2+120a^4+a\right]}^{6}2\left[64+240{\left(\sqrt{1-x}\right)}^4+120{\left(\sqrt{1-x}\right)}^4+{\left(\sqrt{1-x}\right)}^6\right]2\left[64+240\left(1-x\right)+120{\left(1-x\right)}^2+{\left(1-x\right)}^3\right]128+480-480x+240\left(1-2x+x^2\right)+2\left(1-3x+3x^2-x3\right)850-966x+246x^2-2x^3$