If the zeros of polynomial ax cube +3bx square +3cx + d are in AP
prove that2b cube - 3abc + a square d eqal to 0
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Answer:
Let the zeros of the given polynomial be p,q,r. As the roots are in A.P., then it can be assumed as p−k,p,p+k, where k is the common difference.
p−k+p+p+k=−
a
3b
p=−
a
b
And, (p−k)(p)(p+k)=−
a
d
p(p
2
−k
2
)=−
a
d
−
a
b
(p
2
−k
2
)=−
a
d
p
2
−k
2
=
b
d
And, p(p−k)+p(p+k)+(p−k)(p+k)=
a
3c
2p
2
+(p
2
−k
2
)=
a
3c
a
2
2b
2
+
b
d
=
a
3c
a
2
b
2b
3
+a
2
d
=
a
3c
2b
3
−3abc+a
2
d=0
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