\( -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 \), 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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= z^3 + \left(8 - 8\right)z^2 + \left(\frac{64 - 128 + 27}{3}\right)z + \left(\frac{512}{27} - \frac{1536}{27} + \frac{648}{27} - \frac{486}{27}\right)
= z^3 + \left(\frac{-407}{3}\right)z + \left(\frac{512 - 1536 + 648 - 486}{27}\right) = z^3 - \frac{407}{3}z - \frac{1422}{27} = z^3 - \frac{407}{3}z - \frac{474}{9}
= z^3 - \frac{407}{3}z - \frac{158}{3}
\( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)
\( q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} \), so \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)
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