asked by guest
You asked:
Find the zeros of
y 3 + 5 y + 125 {y}^{3} + 5 y + 125 y 3 + 5 y + 125
MathBot Answer:
The real zeros of y 3 + 5 y + 125 y^{3} + 5 y + 125 y 3 + 5 y + 125 y = − 60 3 ( 225 + 50685 ) 2 3 + 2 ⋅ 450 3 6 225 + 50685 3 ≈ − 4.66719320442263 \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*} y = 6 3 225 + 50685 − 3 60 ( 225 + 50685 ) 3 2 + 2 ⋅ 3 450 ≈ − 4.66719320442263
The complex zeros of y 3 + 5 y + 125 y^{3} + 5 y + 125 y 3 + 5 y + 125 y = − 2 ⋅ 450 3 + 60 3 ( 225 + 50685 ) 2 3 + 6 ⋅ 3 6 ⋅ 50 3 i + 20 3 ⋅ 3 5 6 i ( 225 + 50685 ) 2 3 12 225 + 50685 3 ≈ 2.33359660221132 + 4.61920115448078 i y = − 2 ⋅ 450 3 + 60 3 ( 225 + 50685 ) 2 3 − 20 3 ⋅ 3 5 6 i ( 225 + 50685 ) 2 3 − 6 ⋅ 3 6 ⋅ 50 3 i 12 225 + 50685 3 ≈ 2.33359660221132 − 4.61920115448078 i \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*} y y = 12 3 225 + 50685 − 2 ⋅ 3 450 + 3 60 ( 225 + 50685 ) 3 2 + 6 ⋅ 6 3 ⋅ 3 50 i + 3 20 ⋅ 3 6 5 i ( 225 + 50685 ) 3 2 ≈ 2.33359660221132 + 4.61920115448078 i = 12 3 225 + 50685 − 2 ⋅ 3 450 + 3 60 ( 225 + 50685 ) 3 2 − 3 20 ⋅ 3 6 5 i ( 225 + 50685 ) 3 2 − 6 ⋅ 6 3 ⋅ 3 50 i ≈ 2.33359660221132 − 4.61920115448078 i
