Question

Using binomial theorem, expand $\left\{\right(x+y{)}^5+(x-y{)}^5\}$ and hence find the value of $\left\{\right(\sqrt{2}+1{)}^5+(\sqrt{2}-1{)}^5\}$.

Answer 1

$(x+y{)}^5+(x-y{)}^5$

$\Rightarrow {}^5{C}_0{x}^5{y}^0+{}^5{C}_1{x}^{4}y+{}^5{C}_2{x}^3{y}^2+{}^5{C}_3{x}^2{y}^3+{}^5{C}_{4}x{y}^4+{}^5{C}_5{x}^0{y}^5+{}^5{C}_0{x}^5$

$-{}^5{C}_1{x}^{4}y+{}^5{C}_2{x}^3{y}^2+{}^5{C}_3{x}^2{y}^3+{}^5{C}_{4}x{y}^4+{}^5{C}_5{y}^5$

$\Rightarrow 2({}^5{C}_0{x}^5+{}^5{C}_2{x}^3{y}^2+{}^5{C}_{4}x{y}^4)$

$\Rightarrow 2[x^5+10x^3{y}^2+5x{y}^4]$

$\Rightarrow 2x[x^4+10x^2{y}^2+5y^4]$ (equation $1$)

Now, ${(\sqrt{2}+1)}^5+{(\sqrt{2}-1)}^5$

Here, $x=\sqrt{2},y=1$

put in equation $1$ and get

$=2\sqrt{2}\left[\right({\sqrt{2}}^4)+10(\sqrt{2}{)}^2+5]$

$=2\sqrt{2}[4+20+5]$

$=58\sqrt{2}$