If the curves y=x3+ax and y=bx2+c pass through the point (-1,0) and have a common  tangent line at this point, then the value of a-3b+c=

y=x3+ax……………….(1) and y=bx2+c………….(2) Passes through the point (-1,0)⇒0=-1-a,  0=b+c…………(3)a=−1 Also curves have common tangent at the point (-1,0) m1=m2a+3=-2b……… (4) Solving (3) & (4)a=-1 b=-1 c=1  ∴a-3b+c=-1-3(-1)+1=3