What is the proof that a^0=1?
a^0 = a^(1 - 1)
a^0 = [a^1] × [a^(-1)]
a^0 = a × (1/a) [since a^1 = a & a^(-1) = 1/a]
a^0 = a / a
a^0 = 1
Hence proved!
If (1+ax+bx2)4=a0+a1x+a2x2+…+a8x8; a, b, a0, a1…a8ϵR and are such that a0+a1+a2≠0 and ∣∣ ∣∣a0a1a2a1a2a0a2a0a1∣∣ ∣∣=0, then