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Byju's Answer
Standard VIII
Mathematics
Factorisation by Common Factors
Simplify: a2...
Question
Simplify:
a
2
−
3
b
2
−
c
2
−
2
a
b
+
4
b
c
.
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Solution
Now,
a
2
−
3
b
2
−
c
2
−
2
a
b
+
4
b
c
=
(
a
2
−
2
a
b
+
b
2
)
−
4
b
2
−
c
2
+
4
b
c
=
(
a
−
b
)
2
−
(
2
b
−
c
)
2
=
(
a
+
b
−
c
)
(
a
−
b
+
c
)
.
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Similar questions
Q.
Simplify:
2
(
a
b
+
c
d
)
−
a
2
−
b
2
+
c
2
+
d
2
Q.
On simplifying this
(
a
+
b
+
c
)
2
−
a
2
−
b
2
−
c
2
2
(
a
b
+
b
c
+
c
a
)
we get,
Q.
The sides
a
,
b
,
c
of a triangle satisfy the relations
c
2
=
2
a
b
and
a
2
+
c
2
=
3
b
2
. Then the measure of
∠
B
A
C
, in degrees, is
Q.
The sides
a
,
b
,
c
of a triangle satisfy the relations
c
2
=
2
a
b
and
a
2
+
c
2
=
3
b
2
.
Then the measure of
∠
B
A
C
,
in degrees, is
Q.
If
6
a
2
−
3
b
2
−
c
2
+
7
a
b
−
a
c
+
4
b
c
=
0
, then the family of lines
a
x
+
b
y
+
c
=
0
is concurrent at
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