If the zeros of the polynomial 
are in A.P., prove that 
Answers
SOLUTION :
Let, a – d, a, a + d be the zeroes of the polynomial.
Given : f(x)= ax³ +3bx² + 3cx + d
Sum of zeroes = −coefficient of x² / coefficient of x³
α + β + γ = −b/a
α + β + γ = - 3b/a
(a – d) +( a) + (a + d) = - 3b/a
a + a + a -d -d = - 3b/a
3a = - 3b/a
a = - 3b/a × ⅓ = -b/a
a = -b/a
Since, a is the zero of the polynomial f(x),
Therefore, f(a) = 0
a(a²)+ 3b(a)² + 3c(a) + d = 0
On Substituting a = -b/a ,
a(−b/a)³ + 3(b/a)² – 3c(b/a) + d=0
-ab³/a³ + 3b²/a² - 3bc/a + d = 0
-b³/a² + 3b²/a² - 3bc/a + d = 0
(-b³ + 3b² - 3abc + a²d)/a² = 0
2b³ - 3abc + a²d = 0 × a²
2b³ - 3abc + a²d = 0
Hence, it is proved
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