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Byju's Answer
Standard XII
Mathematics
Identity in Binary Operation
a+b+c-c-b-c a...
Question
a
+
b
+
c
-
c
-
b
-
c
a
+
b
+
c
-
a
-
b
-
a
a
+
b
+
c
=
2
a
+
b
b
+
c
c
+
a
Open in App
Solution
Let
LHS
=
∆
=
a
+
b
+
c
-
c
-
b
-
c
a
+
b
+
c
-
a
-
b
-
a
a
+
b
+
c
=
a
-
c
-
b
b
a
+
b
+
c
-
a
c
-
a
a
+
b
+
c
Applying
C
1
→
C
1
+
C
2
+
C
3
=
a
+
b
a
+
b
-
a
+
b
b
+
c
b
+
c
b
+
c
c
-
a
a
+
b
+
c
Applying
R
1
→
R
1
+
R
2
and
R
2
→
R
2
+
R
3
=
a
+
b
b
+
c
1
1
-
1
1
1
1
c
-
a
a
+
b
+
c
Taking
out
common
factor
from
R
1
and
R
2
=
a
+
b
b
+
c
0
0
-
2
1
1
1
c
-
a
a
+
b
+
c
Applying
R
1
→
R
1
-
R
2
=
a
+
b
b
+
c
-
2
-
a
-
c
Expanding
along
R
1
=
2
a
+
b
b
+
c
c
+
a
=
RHS
Hence proved.
Suggest Corrections
2
Similar questions
Q.
If a, b, c are real numbers such that
b
+
c
c
+
a
a
+
b
c
+
a
a
+
b
b
+
c
a
+
b
b
+
c
c
+
a
=
0
, then show that either
a
+
b
+
c
=
0
or
,
a
=
b
=
c
.
Q.
If
a
,
b
and
c
are real numbers and
Δ
=
∣
∣ ∣
∣
b
+
c
c
+
a
a
+
b
c
+
a
a
+
b
b
+
c
a
+
b
b
+
c
c
+
a
∣
∣ ∣
∣
=
0
,
show that either
a
+
b
+
c
=
0
or
a
=
b
=
c
.
Q.
Prove the following identity:
a
(
b
−
c
)
2
(
c
−
a
)
(
a
−
b
)
+
b
(
c
−
a
)
2
(
a
−
b
)
(
b
−
c
)
+
c
(
a
−
b
)
2
(
b
−
c
)
(
c
−
a
)
=
a
+
b
+
c
.
Q.
Prove that
∣
∣ ∣
∣
b
+
c
c
+
a
a
+
b
c
+
a
a
+
b
b
+
c
a
+
b
b
+
c
c
+
a
∣
∣ ∣
∣
=
=
2
(
a
+
b
+
c
)
(
a
b
+
b
c
+
c
a
−
a
2
−
b
2
−
c
2
)
Q.
If
a
,
b
and
c
are real numbers, and
,
Show that either
a
+
b
+
c
= 0 or
a
=
b
=
c
.
,