Recent changes to 387: factor((x^(1/3)+1)^3) failshttps://sourceforge.net/p/maxima/bugs/387/2003-08-19T14:20:11Zfactor((x^(1/3)+1)^3) fails2003-08-19T14:20:11Z2003-08-19T14:20:11ZStavros Macrakishttps://sourceforge.net/u/macrakis/https://sourceforge.neta6220ba8f8ee0ca9b35d823ca352a2400ef7b285factor\(\(x^\(1/3\)+1\)^3\) fails, though factor\(\(x^\(1/3\)+1\)
^2\) works.
The problem is presumably that in the ^2 case, the
powers of x are explicitly of the form n/3, whereas in the
^3 case, \(x^\(1/3\)\)^3 =&gt; x, where the \(1/3\) is not
explicit.
Similarly for \(x^\(1/4\)+1\)^2.
Shouldn't factor normalize the exponents \(like radcan\)?