If xnxn+2x2n1xaaxn+5xa+6x2n+5=0,∀x∈R, where n∈N, then value of a is
n
n - 1
n + 1
none of these
Taking x5 common from last row, we get
x5xnxn+2x2n1xaaxnxa+1x2n=0,∀x∈R
⇒ a+1=n+2 or a=n+1 (as it will make first and third row is identical)