1. If the zeros of f(x)=ax3+3bx2+3cx+ d are in arithmetic progression. Prove that 2b3- 3abc+a2d=0.

Suppose the zeroes of given polynomial in A.P. are l-m; l and l+m
So we have;
x-l-mx-lx-l+m = ax3+3bx2+3cx+dx-lx-l-mx-l+m = ax3+3bx2+3cx+dx-lx2-xl+m-xl-m-l2-m2 = ax3+3bx2+3cx+dx-lx2-lx-mx-lx+mx-l2-m2 = ax3+3bx2+3cx+dx-lx2-2lx-l2-m2 = ax3+3bx2+3cx+dx3-2lx2-xl2-m2-lx2+2l2x+ll2-m2 = ax3+3bx2+3cx+dx3-3lx2-l2x+m2x+2l2x+ll2-m2 = ax3+3bx2+3cx+dx3-3lx2+xl2+m2+ll2-m2 = ax3+3bx2+3cx+dComparing the coefficients of both sides of equation we get;a = 13b = -3lb = -l3c = l2+m2c =  l2+m23And d = ll2-m2Then, 2b3-3abc+a2d= 2-l3-3×1×-l× l2+m23+12× ll2-m2= -2l3+l3+m2l+l3-m2l= 0Hence proved.

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