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Hungary Contests
Kürschák Math Competition
1983 Kurschak Competition
1
1
Part of
1983 Kurschak Competition
Problems
(1)
x^3 + 3y^3 + 9z^3 - 9xyz = 0 in rationals
Source: 1983 Hungary - Kürschák Competition p1
10/10/2022
Let
x
,
y
x, y
x
,
y
and
z
z
z
be rational numbers satisfying
x
3
+
3
y
3
+
9
z
3
−
9
x
y
z
=
0.
x^3 + 3y^3 + 9z^3 - 9xyz = 0.
x
3
+
3
y
3
+
9
z
3
−
9
x
yz
=
0.
Prove that
x
=
y
=
z
=
0
x = y = z = 0
x
=
y
=
z
=
0
.
algebra
rational