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Byju's Answer
Standard IX
Mathematics
Quotient Law
Prove the fol...
Question
Prove the following:
(
x
a
x
b
)
a
2
+
a
b
+
b
2
×
(
x
b
x
c
)
b
2
+
b
c
+
c
2
×
(
x
c
x
a
)
c
2
+
c
a
+
a
2
=
1
Open in App
Solution
(
x
a
x
b
)
a
2
+
a
b
+
b
2
×
(
x
b
x
c
)
b
2
+
b
c
+
c
2
×
(
x
c
x
a
)
c
2
+
a
2
+
c
a
=
x
(
a
−
b
)
(
a
2
+
b
2
+
a
b
)
×
x
(
b
−
c
)
(
b
2
+
b
c
+
c
2
)
×
x
(
c
−
a
)
(
c
2
+
a
2
+
c
a
)
=
x
(
a
3
−
b
3
)
×
x
(
b
3
−
c
3
)
×
x
(
c
3
−
a
3
)
=
x
a
3
−
b
3
+
b
3
−
c
3
+
c
3
−
a
3
=
x
∘
=
1
Suggest Corrections
13
Similar questions
Q.
Simplify :
(
x
a
x
b
)
a
2
+
a
b
+
b
2
×
(
x
b
x
c
)
b
2
+
b
c
+
c
2
×
(
x
c
x
a
)
c
2
+
c
a
+
a
2
Q.
The value of
(
x
a
x
b
)
a
2
+
a
b
+
b
2
⋅
(
x
b
x
c
)
b
2
+
b
c
+
c
2
⋅
(
x
c
x
a
)
c
2
+
c
a
+
b
2
.
Q.
State true or false
Simplified form of
(
x
a
x
b
)
a
2
+
a
b
+
b
2
×
(
x
b
x
c
)
b
2
+
b
c
+
c
2
×
(
x
c
x
a
)
c
2
+
c
a
+
a
2
is 1
Q.
Prove that :
(i)
(
x
a
x
b
)
a
2
+
a
b
+
b
2
×
(
x
b
x
c
)
b
2
+
a
b
+
c
2
×
(
x
c
x
a
)
c
2
+
c
a
+
a
2
=
1
(ii)
(
x
a
x
b
)
c
×
(
x
b
x
c
)
a
×
(
x
c
x
a
)
b
=
1
Q.
(
x
a
x
b
)
a
2
+
a
b
+
b
2
(
x
b
x
c
)
b
2
+
b
c
+
c
2
(
x
c
x
a
)
c
2
+
c
a
+
a
2
equals
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