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2021 BMT
12
BMT 2021 General p12
BMT 2021 General p12
Source:
September 27, 2023
algebra
Problem Statement
Let
a
a
a
,
b
b
b
, and
c
c
c
be the solutions of the equation
x
3
−
3
⋅
202
1
2
x
=
2
⋅
20213.
x^3 - 3 \cdot 2021^2x = 2 \cdot 20213.
x
3
−
3
⋅
202
1
2
x
=
2
⋅
20213.
Compute
1
a
+
1
b
+
1
c
.
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.
a
1
+
b
1
+
c
1
.
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