Question

Find $(x+1{)}^6+(x-1{)}^6$. Hence or otherwise evaluate $(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6$

Answer 1

Using Binomial theorem,
$(x+1{)}^6{=}^6{C}_0{x}^6{+}^6{C}_1{x}^5{+}^6{C}_2{x}^4{+}^6{C}_3{x}^3{+}^6{C}_4{x}^2{+}^6{C}_{5}x{+}^6{C}_6$

$(x-1{)}^6{=}^6{C}_0{x}^6{-}^6{C}_1{x}^5{+}^6{C}_2{x}^4{-}^6{C}_3{x}^3{+}^6{C}_4{x}^2{-}^6{C}_{5}x{+}^6{C}_6$

$\therefore (x+1{)}^6+(x-1{)}^6=2{[}^6{C}_0{x}^6{+}^6{C}_2{x}^4{+}^6{C}_4{x}^2{+}^6{C}_6]=2[x^6+15x^4+15x^2+1]$

By putting $x=\sqrt{2}$ we get,

$(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6=2\left[\right(\sqrt{2}{)}^6+15(\sqrt{2}{)}^4+15(\sqrt{2}{)}^2+1]$

$=2(8+15\times 4+15\times 2+1)$

$=2(8+60+30+1)$

$=2\left(99\right)$

$=198$