\( -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 \), huge positive

February 22, 2026

$\( -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 \), huge positive$

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Hier: \( p = -\frac{407}{3} \), \( q = -\frac{158}{3} \)
\Delta = -4\left(-\frac{407}{3}\right)^3 - 27\left(-\frac{158}{3}\right)^2 = -4\left(-\frac{67432643}{27}\right) - 27\cdot\frac{24964}{9}
Besser: positives Vorzeichen:
\( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)
But \( p \) is large negative denominator: \( p = -\frac{407}{3} \Rightarrow p^3 = -\frac{67432643}{27} \), so \( -4p^3 = -4 \cdot (-\frac{67432643}{27}) = \frac{269730372}{27} \approx 9996788 \)
\( q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} \), so \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)

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