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Byju's Answer
Standard VII
Mathematics
Types of Expressions Based on the Number of Terms
Simplify each...
Question
Simplify each of the following
a
(
b
−
c
)
+
b
(
c
−
a
)
+
c
(
a
−
b
)
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Solution
We have,
a
(
b
−
c
)
+
b
(
c
−
a
)
+
c
(
a
−
b
)
=
a
b
−
a
c
+
b
c
−
a
b
+
a
c
−
b
c
=
0
Hence, this is the answer.
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0
Similar questions
Q.
Simplify the following expression:
(
a
b
+
b
c
)
(
a
b
−
b
c
)
+
(
b
c
+
c
a
)
(
b
c
−
a
c
)
+
(
c
a
+
a
b
)
(
a
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−
a
b
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Q.
Prove the following :
∣
∣ ∣
∣
b
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b
c
′
+
b
′
c
b
′
c
′
c
a
c
a
′
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c
′
a
c
′
a
′
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b
a
b
′
+
d
′
b
d
′
b
′
∣
∣ ∣
∣
=
(
b
c
′
−
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′
c
)
(
c
a
′
−
c
′
a
)
(
a
b
′
−
d
b
)
Q.
Prove that
∣
∣ ∣ ∣
∣
b
c
−
a
2
c
a
−
b
2
a
b
−
c
2
−
b
c
+
c
a
+
a
b
b
c
−
c
a
+
a
b
b
c
+
c
a
−
a
b
(
a
+
b
)
(
a
+
c
)
(
b
+
c
)
(
b
+
a
)
(
c
+
a
)
(
c
+
b
)
∣
∣ ∣ ∣
∣
=
3.
(
b
−
c
)
(
c
−
a
)
(
a
−
b
)
(
a
+
b
+
c
)
(
a
b
+
b
c
+
c
a
)
Q.
Show that :
∣
∣ ∣ ∣
∣
b
c
−
a
2
c
a
−
b
2
a
b
−
c
2
−
b
c
+
c
a
+
c
b
b
c
−
c
a
+
a
b
b
c
+
c
a
−
a
b
(
a
+
b
)
(
a
+
c
)
(
b
+
c
)
(
b
+
a
)
(
c
+
a
)
(
c
+
b
)
∣
∣ ∣ ∣
∣
=
3
(
b
−
c
)
(
c
−
a
)
(
a
−
b
)
(
a
+
b
+
c
)
(
b
c
+
c
a
+
a
b
)
.
Q.
Show that the determinant
∣
∣ ∣ ∣
∣
a
2
+
b
2
+
c
2
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
a
2
+
b
2
+
c
2
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
b
c
+
c
a
+
a
b
a
2
+
b
2
+
c
2
∣
∣ ∣ ∣
∣
is always non-negative.
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