If 1+ax+bx24=a0+a1x+a2x2+⋯+a8x8, where a, b, a0, a1, …,a8∈R such that a0+a1+a2≠0 and a0a1   a2a1a2a0a2a0a1=0, then the value of 5ab is ______ .

Putting x = 0, a0 = 11+ax+bx24=1+ax+bx21+ax+bx21+ax+bx21+ax+bx2Clearly a0 = 1, a1 = coefficient of x = a + a + a + a = 4aa2 = coefficient of x2 = 4b + 6a2Now Δ=−a03+a13+a23−3a0a1a2∵    a0+a1+a2≠0∴     a0=a1=a21 = 4a = 6a2 + 4b⇒ a=14,b=532