Factorize 3p(p + 3)^1/3−8(p + 3)^4/3
You asked:
Factor the expression: \(\displaystyle \frac{3 {\left( p \cdot \left(p + 3\right) \right)}^{1}}{3} - \frac{8 {\left( p + 3 \right)}^{4}}{3}\)
MathBot Answer:
Factorization over the rational numbers:
\[- \frac{\left(p + 3\right) \left(8 p^{3} + 72 p^{2} + 213 p + 216\right)}{3}\]Factorization over the complex numbers:
\[- \frac{8 \left(p + 3\right) \left(p + \left(3 + \frac{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27 \sqrt{322}}{32} + \frac{243}{16}}}{3} + \frac{3}{8 \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27 \sqrt{322}}{32} + \frac{243}{16}}}\right)\right) \left(p + \left(3 + \frac{3}{8 \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27 \sqrt{322}}{32} + \frac{243}{16}}} + \frac{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27 \sqrt{322}}{32} + \frac{243}{16}}}{3}\right)\right) \left(p + \left(\frac{3}{8 \sqrt[3]{\frac{27 \sqrt{322}}{32} + \frac{243}{16}}} + \frac{\sqrt[3]{\frac{27 \sqrt{322}}{32} + \frac{243}{16}}}{3} + 3\right)\right)}{3}\]\(i\) is the imaginary unit, defined as \(i^2 = -1\).