Simplify with the help of binomial theorm.

Answer

N/a

Explanation

`(x+1)^(5)=overset(5)underset(r=0)(Sigma).^(5)C_(r)x^(5-r).1^(r) =overset(5)underset(r=0)(Sigma).^(5)C_(r).x^(5-r)`
`rArr(x+1)^(5)=^(5)C_(0)x^(5)+^(5)C_(1).x^(4)+^(5)C_(2)x^(3)`
`+^(5)C_(3)x^(2)+^(5)C_(4)x+^(5)C_(5).....(i)`
`" and "(x-1)^(5)=overset(5)underset(r=0)(Sigma).^(5)C_(r)x^(5-r)(-1)^(r)`
`=^(5)C_(0)C^(5)+^(5)C_(1)x^(4).(-1)^(1)`
`+^(5)C_(2)x^(3)(-1)^(2)+^(5)C_(3)x^(2).(-1)^(3)`
`+^(5)C_(4)x(-1)^(4)+^(5)C_(5)(-1)^(5)`
`(x-1)^(5)=^(5)C_(0)x^(5)-^(5)C_(1)x^(4)+^(5)C_(2)x^(3)-^(5)C_(3)x^(2)`
`+^(5)C_(4).x-^(5)C_(5)......(2)`
Adding eqs. (1) and (2) we get .
`(x+1)^(5)+(x-1)^(5)=2[^(5)C_(0)x^(5)+^(5)C_(2)x^(3)+^(5)C_(4).x]`
`=2[1.x^(5)+10.x^(3)+5.x]`
`=2[x^(5)+10x^(3)+5x).`