27·x3 - y3 - 13·x·y/z = 0
3·x2·z - 4·x·y + 3/z = 0
3·x·z - y·z = 1 --> z = 1/(3·x - y) in I und II einsetzen
27·x3 - y3 - 13·x·y·(3·x - y) = 0 --> (x - y)·(y - 3·x)·(y - 9·x) = 0 --> y = 9·x ∨ y = 3·x ∨ y = x
3·x2/(3·x - y) - 4·x·y + 3·(3·x - y) = 0
3·x2 - 4·x·y·(3·x - y) + 3·(3·x - y)·(3·x - y) = 0
- 12·x2·y + 30·x2 + 4·x·y2 - 18·x·y + 3·y2 = 0
Hier die Lösungen von II einsetzen
y = x
- 12·x2·x + 30·x2 + 4·x·x2 - 18·x·x + 3·x2 = 0 --> x = 15/8 (∨ x = 0)
y = 3·x
- 12·x2·(3·x) + 30·x2 + 4·x·(3·x)2 - 18·x·(3·x) + 3·(3·x)2 = 0 → (x = 0)
y = 9·x
- 12·x2·(9·x) + 30·x2 + 4·x·(9·x)2 - 18·x·(9·x) + 3·(9·x)2 = 0 --> x = - 37/72 (∨ x = 0)
Damit kann man jetzt auch y und z ausrechnen.