Question

if the zeroes of p(x)=ax³+3bx²+3cx+dare in AP. Prove that 2b³-3abc+a²d=0.

Answer 1

We may then write the polynomial as 
(x- (m – n))(x – m)(x – (m + n)) = 0 

Hence 
f(x) = a(x – m + n)(x – m)(x – m – n) 

Comparing coefficients with ax^3 +3bx^2 +3cx +d 

For x^2 
3b = -3am so that b = -am 

For x 
3c = 3am^2 – an^2 so that 3ac = 3b^2 – a^2n^2
and then a^2n^2 = 3b^2 – 3ac 

For units 
d = -am^3 + amn^2 so that a^2d = b ^3 – a^2bn^2 

and then 
a^2d = b ^3 - b(3b^2 – 3ac) 
a^2d = b ^3 - 3b^3 + 3abc 
which means 
2b^3 - 3abc + a^2d = 0 
as required