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Let p, q and r be the zeroes of the polynomial ax³+3bx²+3cx -d

Since the zeroes are in AP,  2q = p + r ..........(1)

Sum of the zeroes of a cubic polynomial , p + q + r = -3b/a ................(2)

Sumof zeroes taking two at a time, pq + qr + rp = 3c/a .................(3)

Product of zeroes, pqr = d/a  .....................(4)

From (1) and (2), we get

3q = -3b/ a

⇒ q = -b/a  .....................(5)

Now from equation (4)

pr = d/ aq  = -d/b ....................(6)

Now, from equation (3)

 pq + qr + rp = 3c/a

⇒ q (p+r) + rp = 3c/a

⇒ 2(-b/a)² - (d/b) = 3c/a

Solving the above you will get the answer as 2b³-3abc - a²d = 0