how to factorize 27y-9y to the power of 3 +3y to the power of 4
You asked:
Factor the expression: \(\displaystyle 27 y - 9 \cdot {y}^{3} + 3 \cdot {y}^{4}\)
MathBot Answer:
Factorization over the rational numbers:
\[3 y \left(y^{3} - 3 y^{2} + 9\right)\]Factorization over the complex numbers:
\[3 y \left(y + \left(-1 + \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{5}}{2} + \frac{7}{2}} + \frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{5}}{2} + \frac{7}{2}}}\right)\right) \left(y + \left(-1 + \frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{5}}{2} + \frac{7}{2}}} + \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{3 \sqrt{5}}{2} + \frac{7}{2}}\right)\right) \left(y + \left(-1 + \frac{1}{\sqrt[3]{\frac{3 \sqrt{5}}{2} + \frac{7}{2}}} + \sqrt[3]{\frac{3 \sqrt{5}}{2} + \frac{7}{2}}\right)\right)\]\(i\) is the imaginary unit, defined as \(i^2 = -1\).