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Byju's Answer
Standard XII
Mathematics
Integration by Partial Fractions
Evaluate x ...
Question
Evaluate
(
x
+
α
)
(
x
+
β
)
(
α
−
γ
)
(
β
−
γ
)
+
(
x
+
β
)
(
x
+
γ
)
(
β
−
α
)
(
γ
−
α
)
+
(
x
+
α
)
(
x
+
γ
)
(
γ
−
β
)
(
α
−
β
)
=
?
Open in App
Solution
Now,
(
x
+
α
)
(
x
+
β
)
(
α
−
γ
)
(
β
−
γ
)
+
(
x
+
β
)
(
x
+
γ
)
(
β
−
α
)
(
γ
−
α
)
+
(
x
+
α
)
(
x
+
γ
)
(
γ
−
β
)
(
α
−
β
)
x
2
+
x
(
α
+
β
)
+
α
.
β
(
α
−
γ
)
(
β
−
γ
)
+
x
2
+
x
(
β
+
γ
)
+
β
.
γ
(
β
−
α
)
(
γ
−
α
)
+
x
2
+
x
(
α
+
γ
)
+
α
.
γ
(
γ
−
β
)
(
α
−
β
)
=
{
x
2
[
(
β
−
α
)
+
(
γ
−
β
)
+
(
α
−
γ
)
]
+
x
[
(
β
2
−
α
2
)
+
(
γ
2
−
α
2
)
+
(
α
2
−
γ
2
)
]
+
[
α
.
β
(
β
−
α
)
+
β
.
γ
(
−
β
+
γ
)
+
γ
.
α
(
α
−
γ
)
]
}
(
α
−
β
)
(
β
−
γ
)
(
γ
−
α
)
)
=
{
[
α
.
β
(
β
−
α
)
+
β
.
γ
(
−
β
+
γ
)
+
γ
.
α
(
α
−
γ
)
]
}
(
α
−
β
)
(
β
−
γ
)
(
γ
−
α
)
)
Suggest Corrections
0
Similar questions
Q.
If
α
,
β
,
γ
are roots of the equation
x
3
+
2
x
−
3
=
0
, then the equation whose roots are
(
α
−
β
)
(
α
−
γ
)
,
(
β
−
γ
)
(
β
−
α
)
,
(
γ
−
α
)
(
γ
−
β
)
is:
Q.
If
∣
∣ ∣ ∣
∣
(
β
+
γ
−
α
−
δ
)
4
(
β
+
γ
−
α
−
δ
)
2
1
(
γ
+
α
−
β
−
δ
)
4
(
γ
+
α
−
β
−
δ
)
2
1
(
α
+
β
−
γ
−
δ
)
4
(
α
+
β
−
γ
−
δ
)
2
1
∣
∣ ∣ ∣
∣
=
−
K
(
α
−
β
)
(
α
−
γ
)
(
β
−
γ
)
(
β
−
δ
)
(
γ
−
δ
)
Then the value of
K
is:
Q.
Prove that
∣
∣ ∣ ∣
∣
(
β
+
γ
−
α
−
δ
)
4
(
β
+
γ
−
α
−
δ
)
2
1
(
γ
+
α
−
β
−
δ
)
4
(
γ
+
α
−
β
−
δ
)
2
1
(
α
+
β
−
γ
−
δ
)
4
(
α
+
β
−
γ
−
δ
)
2
1
∣
∣ ∣ ∣
∣
=
−
64
(
α
−
β
)
(
α
−
γ
)
(
α
−
δ
)
(
β
−
γ
)
(
β
−
δ
)
(
γ
−
δ
)
Q.
If
∣
∣ ∣ ∣
∣
(
β
+
γ
−
α
−
δ
)
4
(
β
+
γ
−
α
−
δ
)
2
1
(
γ
+
α
−
β
−
δ
)
4
(
γ
+
α
−
β
−
δ
)
2
1
(
α
+
β
−
γ
−
δ
)
4
(
α
+
β
−
γ
−
δ
)
2
1
∣
∣ ∣ ∣
∣
=
−
k
(
α
−
β
)
(
α
−
γ
)
(
α
−
δ
)
(
β
−
γ
)
(
β
−
δ
)
(
γ
−
δ
)
, then the value of
(
k
)
1
2
Q.
If
α
,
β
,
γ
,
δ
are in
A
.
P
.
and
∫
2
0
f
(
x
)
d
x
=
−
4
where
f
(
x
)
=
∣
∣ ∣ ∣
∣
x
+
α
x
+
β
x
+
α
−
γ
x
+
β
x
+
γ
x
−
1
x
+
γ
x
+
δ
x
−
β
+
δ
∣
∣ ∣ ∣
∣
then the common difference
d
is:
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