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Standard IX
Mathematics
Question
Solve
9
1
3
×
27
−
1
2
3
1
6
×
3
−
2
3
Open in App
Solution
Verified by Toppr
9
1
3
×
27
−
1
2
3
1
6
×
3
−
2
3
=
(
3
2
)
1
3
×
(
3
3
)
−
1
2
3
1
6
×
3
−
2
3
=
3
2
3
×
3
−
3
2
3
1
6
×
3
−
2
3
(
a
s
(
a
m
)
n
=
a
m
n
)
=
3
2
3
−
3
2
3
1
6
−
2
3
(
a
s
a
m
×
a
n
=
a
m
+
n
)
=
3
−
5
6
3
−
1
2
=
3
−
5
6
−
(
−
1
2
)
(
a
s
a
m
a
n
=
a
m
−
n
)
=
3
1
2
−
5
6
=
3
−
1
3
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