Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If f(x)=a+b x+c x2 and α, β, γ are roots of the equation x3=1, then |a&b&c b&c&a c&a&b| is equal to
Q. If
f
(
x
)
=
a
+
b
x
+
c
x
2
and
α
,
β
,
γ
are roots of the equation
x
3
=
1
, then
∣
∣
a
b
c
b
c
a
c
a
b
∣
∣
is equal to
2099
123
Determinants
Report Error
A
f
(
α
)
+
f
(
β
)
+
f
(
γ
)
B
f
(
α
)
f
(
β
)
+
f
(
β
)
f
(
γ
)
+
f
(
γ
)
f
(
α
)
C
f
(
α
)
f
(
β
)
f
(
γ
)
D
−
f
(
α
)
f
(
β
)
f
(
γ
)
Solution:
∣
∣
a
b
c
b
c
a
c
a
b
∣
∣
=
−
(
a
3
+
b
3
+
c
3
−
3
ab
c
)
=
−
(
a
+
b
+
c
)
(
a
+
b
ω
2
+
c
ω
)
(
a
+
bω
+
c
ω
2
)
=
−
f
(
α
)
f
(
β
)
f
(
γ
)
[
∵
α
=
1
,
β
=
ω
,
γ
=
ω
2
]