x^3 + x - 80 = 0

asked by guest
on Jan 12, 2025 at 4:06 am



You asked:

Solve the equation \({x}^{3} + x - 80 = 0\) for the variable \(x\).

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}x &= \sqrt[3]{40 + \frac{\sqrt{129603}}{9}} - \frac{1}{3 \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}} \approx 4.231518\\x &= - \frac{\sqrt[3]{40 + \frac{\sqrt{129603}}{9}}}{2} + \frac{1}{6 \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}} + i \left(\frac{\sqrt{3}}{6 \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}} + \frac{\sqrt{3} \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}}{2}\right) \approx -2.115759 + 3.798593 i\\x &= - \frac{\sqrt[3]{40 + \frac{\sqrt{129603}}{9}}}{2} + \frac{1}{6 \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}} + i \left(- \frac{\sqrt{3} \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}}{2} - \frac{\sqrt{3}}{6 \sqrt[3]{40 + \frac{\sqrt{129603}}{9}}}\right) \approx -2.115759 -3.798593 i\end{aligned}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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