Question

The value of $(\sqrt{2}+1{)}^6+(\sqrt{2}-1{)}^6$ is

Answer 1

We know that
$(a+b{)}^6={}^6{C}_0{a}^6+{}^6{C}_1{a}^{5}b+{}^6{C}_2{a}^4{b}^2+\cdots +{}^6{C}_6{a}^0{b}^6(a-b{)}^6={}^6{C}_0{a}^6-{}^6{C}_1{a}^{5}b+{}^6{C}_2{a}^4{b}^2-\cdots +{}^6{C}_6{a}^0{b}^6$
Adding both we get
$(a+b{)}^6+(a-b{)}^6=2[a^6+{}^6{C}_2{a}^4{b}^2+{}^6{C}_4{a}^2{b}^4+b^6]$

Here, $a=\sqrt{2},b=1$
$=2\left[\right(\sqrt{2}{)}^6+{}^6{C}_2(\sqrt{2}{)}^4{1}^2+{}^6{C}_4(\sqrt{2}{)}^2{1}^4+1^6]$
$=2[8+15\times 4+15\times 2+1]$
$=198$