Consider the position vector $OA=(t^{2},1-t,1-t^{2}),t\geq 0$. What is the value of $t$ that makes the distance to the origin a minimum?
My reasoning is to find the derivative of the Euclidean distance and set it to zero. I get $4t^{3}=t+1$. Numerically this gives $t=0.706$. Is there a way to solve $4t^{3}=t+1$ analytically instead? (say to obtain the root of some number)
$$\boxed{ t = \left( \frac{1}{8} - \frac{\sqrt{78}}{72} \right)^\tfrac{1}{3} + \left( \frac{1}{8} + \frac{\sqrt{78}}{72} \right)^\tfrac{1}{3} }$$
How? I entered $A t^3 -B t = 1$ to a CAS system (like Wolfram Alpha) and it solved the cubic equation. If you want to know more about the roots of a cubic polynomial, look at equations (71), (72) and (73) in this write-up from Wolfram.