y^3 + 5y + 125

find roots

asked by guest
on Nov 17, 2024 at 1:22 am



You asked:

Find the zeros of \({y}^{3} + 5 y + 125\)

MathBot Answer:

The real zeros of \(y^{3} + 5 y + 125\) are: $$\begin{align*}y &= \frac{- \sqrt[3]{60} \left(225 + \sqrt{50685}\right)^{\frac{2}{3}} + 2 \cdot \sqrt[3]{450}}{6 \sqrt[3]{225 + \sqrt{50685}}} \approx -4.66719320442263\end{align*}$$

The complex zeros of \(y^{3} + 5 y + 125\) are: $$\begin{align*}y &= \frac{- 2 \cdot \sqrt[3]{450} + \sqrt[3]{60} \left(225 + \sqrt{50685}\right)^{\frac{2}{3}} + 6 \cdot \sqrt[6]{3} \cdot \sqrt[3]{50} i + \sqrt[3]{20} \cdot 3^{\frac{5}{6}} i \left(225 + \sqrt{50685}\right)^{\frac{2}{3}}}{12 \sqrt[3]{225 + \sqrt{50685}}} \approx 2.33359660221132 + 4.61920115448078 i\\y &= \frac{- 2 \cdot \sqrt[3]{450} + \sqrt[3]{60} \left(225 + \sqrt{50685}\right)^{\frac{2}{3}} - \sqrt[3]{20} \cdot 3^{\frac{5}{6}} i \left(225 + \sqrt{50685}\right)^{\frac{2}{3}} - 6 \cdot \sqrt[6]{3} \cdot \sqrt[3]{50} i}{12 \sqrt[3]{225 + \sqrt{50685}}} \approx 2.33359660221132 - 4.61920115448078 i\end{align*}$$

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