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Byju's Answer
Standard XII
Mathematics
In Radius
Prove that:1 ...
Question
Prove that :
1
a
2
+
b
c
a
3
1
b
2
+
c
a
b
3
1
c
2
+
a
b
c
3
=
-
a
-
b
b
-
c
c
-
a
a
2
+
b
2
+
c
2
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Solution
Let
LHS
=
Δ
=
1
a
2
+
bc
a
3
1
b
2
+
ca
b
3
1
c
2
+
ab
c
3
⇒
Δ
=
0
a
2
+
bc
-
b
2
+
ca
a
3
-
b
3
0
b
2
+
ca
-
c
2
+
ab
b
3
-
c
3
1
c
2
+
ab
c
Applying
R
1
→
R
1
-
R
2
and
R
2
→
R
2
-
R
3
=
0
a
2
-
b
2
-
ca
+
bc
a
3
-
b
3
0
b
2
-
c
2
-
ab
+
ca
b
3
-
c
3
1
c
2
+
ab
c
3
=
0
a
-
b
a
+
b
-
c
a
-
b
a
2
+
ab
+
b
2
0
b
-
c
b
+
c
-
a
b
-
c
b
2
+
bc
+
a
2
1
c
2
+
ab
c
3
=
a
-
b
b
-
c
0
a
+
b
-
c
a
2
+
ab
+
b
2
0
b
+
c
-
a
b
2
+
bc
+
c
2
1
c
2
+
ab
c
3
Taking
out
a
-
b
common
from
R
1
and
b
-
c
from
R
2
=
a
-
b
b
-
c
0
a
+
b
-
c
a
2
+
ab
+
b
2
0
b
+
c
-
a
-
a
+
b
-
c
b
2
+
bc
+
c
2
-
a
2
+
ab
+
b
2
1
c
2
+
ab
c
3
Applying
R
2
→
R
2
-
R
1
=
a
-
b
b
-
c
0
a
+
b
-
c
a
2
+
ab
+
b
2
0
2
c
-
a
b
c
-
a
+
c
2
-
a
2
1
c
2
+
ab
c
3
=
a
-
b
b
-
c
c
-
a
0
a
+
b
-
c
a
2
+
ab
+
b
2
0
2
a
+
b
+
c
1
c
2
+
ab
c
3
=
a
-
b
b
-
c
c
-
a
×
1
×
a
+
b
-
c
a
2
+
ab
+
b
2
2
a
+
b
+
c
Expanding
along
C
1
=
a
-
b
b
-
c
c
-
a
×
a
+
b
2
-
c
2
-
2
a
2
+
2
ab
+
2
b
2
=
a
-
b
b
-
c
c
-
a
a
+
b
2
-
c
2
-
a
+
b
2
-
a
2
+
b
2
=
-
a
-
b
b
-
c
c
-
a
a
2
+
b
2
+
c
2
=
RHS
Hence proved.
Suggest Corrections
0
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Prove that
∣
∣ ∣
∣
b
+
c
c
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a
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Q.
Prove that :
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Q.
Using properties of determinants, prove that :
∣
∣ ∣
∣
a
b
−
c
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+
b
a
+
c
b
c
−
a
a
−
b
b
+
a
c
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
)
Q.
Prove the following :
∣
∣ ∣
∣
a
b
−
c
c
+
b
a
+
c
b
c
−
a
a
−
b
b
+
a
c
∣
∣ ∣
∣
=
(
a
+
b
+
c
)
(
a
2
+
b
2
+
c
2
)
.
Q.
Prove that :
a
2
a
2
-
b
-
c
2
b
c
b
2
b
2
-
c
-
a
2
c
a
c
2
c
2
-
a
-
b
2
a
b
=
a
-
b
b
-
c
c
-
a
a
+
b
+
c
a
2
+
b
2
+
c
2
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