f(x-2)= x^3 then f(x) =

asked by guest
on Nov 16, 2024 at 1:15 pm



MathBot Answer:

f(x)∈{f(−f27f+−108f3+2916f223−27f+−108f3+2916f2233),f(−f(−12−3i2)27f+−108f3+2916f223−(−12−3i2)27f+−108f3+2916f2233),f(−f(−12+3i2)27f+−108f3+2916f223−(−12+3i2)27f+−108f3+2916f2233)}f{\left(x \right)} \in \left\{f{\left(- \frac{f}{\sqrt[3]{27 f + \frac{\sqrt{- 108 f^{3} + 2916 f^{2}}}{2}}} - \frac{\sqrt[3]{27 f + \frac{\sqrt{- 108 f^{3} + 2916 f^{2}}}{2}}}{3} \right)}, f{\left(- \frac{f}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 f + \frac{\sqrt{- 108 f^{3} + 2916 f^{2}}}{2}}} - \frac{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 f + \frac{\sqrt{- 108 f^{3} + 2916 f^{2}}}{2}}}{3} \right)}, f{\left(- \frac{f}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 f + \frac{\sqrt{- 108 f^{3} + 2916 f^{2}}}{2}}} - \frac{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 f + \frac{\sqrt{- 108 f^{3} + 2916 f^{2}}}{2}}}{3} \right)}\right\}

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