Question

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

Answer 1

We have,

$(x+1{)}^6+(x-1{)}^6$

${[}^6{C}_0{x}^6{+}^6{C}_1{x}^5{+}^6{C}_2{x}^4{+}^6{C}_3{x}^3{+}^6{C}_3{x}^3{+}^6{C}_4{x}^2{+}^6{C}_5{x}^1{+}^6{C}_5{x}^1{+}^6{C}_6{x}^0]+{[}^6{C}_0{x}^6(-1{)}^0{+}^6{C}_1{x}^5(-1{)}^1{+}^6{C}_2{x}^4(-1{)}^2{+}^6{C}_3{x}^3(-1{)}^3{+}^6{C}_4{x}^2(-1{)}^4{+}^6{C}_5{x}^1(-1{)}^5{+}^6{C}_6{x}^0(-1{)}^6]$

${[}^6{C}_0{x}^6{+}^6{C}_1{x}^5{+}^6{C}_2{x}^4{+}^6{C}_3{x}^3{+}^6{C}_4{x}^2{+}^6{C}_4{x}^2{+}^6{C}_{5}x{+}^6{C}_6{+}^6{C}_0{x}^6{-}^6{C}_1{x}^5{+}^6{C}_1{x}^5{+}^6{C}_2{x}^4{}^{}{-}^6{C}_3{x}^3{+}^6{C}_4{x}^2{-}^6{C}_{5}x{+}^6{C}_6]$

$=2{[}^6{C}_0{x}^6{+}^6{C}_2{x}^4{+}^6{C}_4{x}^2{+}^6{C}_6]$

$=2[x^6+15x^4+15x^2=1]$

$\therefore (x+1{)}^6+(x-1{)}^6=2[x^6+15x^4+15x^2+1]....\left(i\right)$

Putting $x=\sqrt{2}$ in equation (i), we get

$(x+1{)}^6+(x-1{)}^6$

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

=2[8+60+30+1]=2[99]=198

$\therefore (x+1{)}^6+(x-1{)}^6=198$