help7

x^3-6x=6  ...   x=2^(2/3)+2^(1/3)

$\boxed{~ \begin{array}{rcll} x^3 + px + q &=& 0 \\ x &=& \sqrt[3]{ -\frac{q}{2} + \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } + \sqrt[3]{ -\frac{q}{2} - \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } \\ \text{If } \frac{q^2}{4} + \frac{p^3}{27} \ge 0 \end{array} ~}$

$\begin{array}{rcll} x^3-6x &=& 6 \\ x^3-6x -6 &=& 0 \\\\ x^3 + px + q &=& 0 \quad | \quad p = -6 = \quad q = -6 \\ x &=& \sqrt[3]{ -\frac{q}{2} + \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } + \sqrt[3]{ -\frac{q}{2} - \sqrt{ \frac{q^2}{4} + \frac{p^3}{27} } } \\ x &=& \sqrt[3]{ -\frac{-6}{2} + \sqrt{ \frac{(-6)^2}{4} + \frac{(-6)^3}{27} } } + \sqrt[3]{ -\frac{-6}{2} - \sqrt{ \frac{(-6)^2}{4} + \frac{(-6)^3}{27} } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 9 - \frac{(6)^3}{27} } } + \sqrt[3]{ 3 - \sqrt{ 9 - \frac{(6)^3}{27} } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 9 - \frac{8\cdot 27}{27} } } + \sqrt[3]{ 3 - \sqrt{ 9 - \frac{8\cdot 27}{27} } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 9 - 8 } } + \sqrt[3]{ 3 - \sqrt{ 9 - 8 } } \\ x &=& \sqrt[3]{ 3 + \sqrt{ 1 } } + \sqrt[3]{ 3 - \sqrt{ 1 } } \\ x &=& \sqrt[3]{ 3 + 1 } + \sqrt[3]{ 3 - 1 } \\ x &=& \sqrt[3]{ 4 } + \sqrt[3]{ 2 } \\ x &=& \sqrt[3]{ 2^2 } + \sqrt[3]{ 2 } \\ x &=& 2^{\frac23} + 2^{\frac13} \\ \end{array}$

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