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Byju's Answer
Standard XII
Mathematics
Special Determinants
= 1+a2+a4 1...
Question
â–³
=
∣
∣ ∣ ∣
∣
1
+
a
2
+
a
4
1
+
a
b
+
a
2
b
2
1
+
a
c
+
a
2
c
2
1
+
a
b
+
a
2
b
2
1
+
b
2
+
b
4
1
+
b
c
+
b
2
c
2
1
+
a
c
+
a
2
c
2
1
+
b
c
+
b
2
c
2
1
+
c
2
+
c
4
∣
∣ ∣ ∣
∣
A
(
a
−
b
)
2
(
b
−
c
)
2
(
c
−
a
)
2
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B
2
(
a
−
b
)
(
b
−
c
)
(
c
−
a
)
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C
4
(
a
−
b
)
(
b
−
c
)
(
c
−
a
)
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D
None of these
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Solution
The correct option is
A
None of these
△
=
∣
∣ ∣ ∣
∣
1
+
a
2
+
a
4
1
+
a
b
+
a
2
b
2
1
+
a
c
+
a
2
c
2
1
+
a
b
+
a
2
b
2
1
+
b
2
+
b
4
1
+
b
c
+
b
2
c
2
1
+
a
c
+
a
2
c
2
1
+
b
c
+
b
2
c
2
1
+
c
2
+
c
4
∣
∣ ∣ ∣
∣
Put
⇒
a
=
1
;
b
=
2
;
c
=
3
△
=
∣
∣ ∣
∣
3
7
1
7
3
3
1
3
3
∣
∣ ∣
∣
△
=
(
9
−
9
)
−
7
(
21
−
3
)
+
1
(
21
−
3
)
=
0
−
7
(
18
)
+
18
=
18
(
−
7
+
1
)
=
18
(
−
6
)
△
=
−
108
∴
None of these.
Suggest Corrections
0
Similar questions
Q.
Prove that
∣
∣ ∣ ∣
∣
1
+
a
2
+
a
4
1
+
a
b
+
a
2
b
2
1
+
a
c
+
a
2
c
2
1
+
a
b
+
a
2
b
2
1
+
b
2
+
b
4
1
+
b
c
+
b
2
c
2
1
+
a
c
+
a
2
c
2
1
+
b
c
+
b
2
c
2
1
+
c
2
+
c
4
∣
∣ ∣ ∣
∣
=
(
a
−
b
)
2
(
b
−
c
)
2
(
c
−
a
)
2
Q.
(
a
2
-
b
2
)
3
+
(
b
2
-
c
2
)
3
+
(
c
2
-
a
2
)
3
(
a
-
b
)
3
+
(
b
-
c
)
3
+
(
c
-
a
)
3
=
(a) 3(a + b) ( b+ c) (c + a)
(b) 3(a − b) (b − c) (c − a)
(c) (a − b) (b − c) (c − a)
(d) none of these