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UPSC NDA I 2015 Mathematics Solved Paper
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© examsnet.com
Question : 13
Total: 120
If
1
,
ω
,
ω
2
are the cube roots of unity, then the value of
(
1
+
ω
)
(
1
+
ω
2
)
(
1
+
ω
4
)
(
1
+
ω
8
)
is
[NDA I 2015]
−
1
0
1
2
Validate
Solution:
We know the following properties of the cube roots of unity
1
+
ω
+
ω
2
=
0
ω
3
=
1
Substituting in the above equation, we get
(
1
+
ω
)
(
1
+
ω
2
)
(
1
+
ω
4
)
(
1
+
ω
8
)
=
(
−
ω
2
)
(
−
ω
)
(
1
+
(
ω
.
ω
3
)
)
(
1
+
(
ω
2.
ω
6
)
)
=
ω
3
(
1
+
ω
)
(
1
+
ω
2
)
=
1.
(
−
ω
2
)
(
−
ω
)
=
ω
3
=
1
© examsnet.com
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