The equation $\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{r} + \sqrt[3]{s} + \sqrt[3]{t}$ can be solved with rational numbers $r, s, t$. The values are determined as $r = \frac{1}{9}$, $s = -\frac{2}{9}$, and $t = \frac{4}{9}$. This solution was verified by cubing both sides of the equation, confirming the correctness of the values derived through trial and error.
PREREQUISITESMathematicians, students studying algebra, and anyone interested in solving cubic equations involving rational numbers.
jacks said:Calculate number $r,s,t$ in $\displaystyle \sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{r} + \sqrt[3]{s} + \sqrt[3]{t}$
where $r,s,t \in Q$
I found that $ \sqrt[3]{\mathstrut\sqrt[3]{2} - 1} = \sqrt[3]{\frac19} + \sqrt[3]{-\frac29} + \sqrt[3]{\frac49}$. I did this more or less by trial and error, so the method is not very revealing. But you can verify that it is correct by cubing both sides.jacks said:Calculate number $r,s,t$ in $\displaystyle \sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{r} + \sqrt[3]{s} + \sqrt[3]{t}$
where $r,s,t \in Q$