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Mathematics
Mathematics
Problem
If
(
1
+
a
x
+
b
x
2
)
4
=
a
0
+
a
1
x
+
a
2
x
2
+
⋯
+
a
8
x
8
(1 + ax + bx^2)^4 = a_0 + a_1 x + a_2 x^2 + \dots + a_8 x^8
(
1
+
a
x
+
b
x
2
)
4
=
a
0
+
a
1
x
+
a
2
x
2
+
⋯
+
a
8
x
8
, where
a
,
b
,
a
0
,
a
1
,
…
,
a
8
∈
R
a, b, a_0, a_1, \dots, a_8 \in \mathbb{R}
a
,
b
,
a
0
,
a
1
,
…
,
a
8
∈
R
such that
a
0
+
a
1
+
a
2
≠
0
a_0 + a_1 + a_2 \neq 0
a
0
+
a
1
+
a
2
=
0
and
∣
a
0
a
1
a
2
a
1
a
2
a
0
a
2
a
0
a
1
∣
=
0
\begin{vmatrix} a_0 & a_1 & a_2 \\ a_1 & a_2 & a_0 \\ a_2 & a_0 & a_1 \end{vmatrix} = 0
a
0
a
1
a
2
a
1
a
2
a
0
a
2
a
0
a
1
=
0
, then a.
a
=
3
4
,
b
=
5
8
a = \frac{3}{4}, b = \frac{5}{8}
a
=
4
3
,
b
=
8
5
b.
a
=
1
4
,
b
=
5
32
a = \frac{1}{4}, b = \frac{5}{32}
a
=
4
1
,
b
=
32
5
c.
a
=
1
,
b
=
2
3
a = 1, b = \frac{2}{3}
a
=
1
,
b
=
3
2
d. None of these
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