Home / $ q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} $, so $ 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 $

February 22, 2026

$q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} \), so \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892$

= z^3 - \frac{407}{3}z - \frac{158}{3}
$ -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 $, huge positive
$ 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 $
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