Axcube+bxsquare +cx+d is divisibleby axsquare+c then abcd are in which progression
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Answer:
c
3
a
=
b
3
d
Explanation:
If
α
,
β
and
γ
are the three roots then we must have
a
x
3
+
b
x
2
+
c
x
+
d
≡
a
(
x
−
α
)
(
x
−
β
)
(
x
−
γ
)
comparing coefficients of various powers of
x
on both sides leads to
α
+
β
+
γ
=
−
b
a
[
1
]
α
β
+
β
γ
+
γ
α
=
+
c
a
[
2
]
α
β
γ
=
−
d
a
[
3
]
In this problem, the three roots are in GP, so that
β
=
α
r
and gamma = alpha r^2#. Substituting this in [1]. [2] and [3] gives
α
(
1
+
r
+
r
2
)
=
−
b
a
[
1
a
]
α
2
(
r
+
r
2
+
r
3
)
=
+
c
a
[
2
a
]
α
3
r
3
=
−
d
a
[
3
a
]
Divideing [2a] by [1a] leads to
α
r
=
−
c
b
and substituting this in [3a] gives
(
−
c
b
)
3
=
−
d
a
⇒
−
c
3
b
3
=
−
d
a
⇒
c
3
a
=
b
3
d
Answer:
c
3
a
=
b
3
d
Explanation:
If
α
,
β
and
γ
are the three roots then we must have
a
x
3
+
b
x
2
+
c
x
+
d
≡
a
(
x
−
α
)
(
x
−
β
)
(
x
−
γ
)
comparing coefficients of various powers of
x
on both sides leads to
α
+
β
+
γ
=
−
b
a
[
1
]
α
β
+
β
γ
+
γ
α
=
+
c
a
[
2
]
α
β
γ
=
−
d
a
[
3
]
In this problem, the three roots are in GP, so that
β
=
α
r
and gamma = alpha r^2#. Substituting this in [1]. [2] and [3] gives
α
(
1
+
r
+
r
2
)
=
−
b
a
[
1
a
]
α
2
(
r
+
r
2
+
r
3
)
=
+
c
a
[
2
a
]
α
3
r
3
=
−
d
a
[
3
a
]
Divideing [2a] by [1a] leads to
α
r
=
−
c
b
and substituting this in [3a] gives
(
−
c
b
)
3
=
−
d
a
⇒
−
c
3
b
3
=
−
d
a
⇒
c
3
a
=
b
3
d
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