Strange result of symbolically solving cubic equation?

I am trying to use solve() to solve the cubic equation w*t^3 - 3*w*t^2 + 3*w*t - r = 0 for t, where 0<=t<=1, using the following code:

clear;
syms w t r;
expr = w*t^3 - 3*w*t^2 + 3*w*t - r;

% symbolic roots of t
tval = solve(expr, t);

% validate t
stval = subs(tval, {w}, {20});
ft = subs(stval, r, 17.5);

I could get the correct result of t = 0.5, just as ft shows. However, if I replace w by b-a, like the following code,

clear;
syms a b w t r;
w = b-a;
expr = w*t^3 - 3*w*t^2 + 3*w*t - r;

% symbolic roots of t
tval = solve(expr, t);

% validate t
stval = subs(tval, {a b}, {0 20});
ft = subs(stval, r, 17.5);

the result of t becomes incorrect ( ft=1 ).

It seems that the tval in later code has items like (d^2)^(1/2), where d is negative, so the item would be simplified as -d. But in the earlier code, these items are simplified to d which is correct.

Does anybody have any ideas about it?

Thanks so much!

[Walter Roberson]

Maple generates identical results for the two situations.

Answer 1

[Andrew Newell]

I don't know why this is happening. I tried the same equations in Maple and got the right answer both ways, so maybe it is a bug in the Matlab implementation. The question you're probably thinking is: How can I be sure I'm getting the right answer? I can suggest two approaches to improve your chances. First, try substituting the solutions back into the equation:

simplify(subs(expr,t,tval(1)))      % also tval(2), tval(3)

This gives zero for your first example but not for your second. My other suggestion is to substitute first, then solve:

clear
syms w t r
expr = subs(w*t^3 - 3*w*t^2 + 3*w*t - r,{w r},{20 17.5});
tval = solve(expr,t)

tval =

                 1/2
 5/4 - (3^(1/2)*i)/4
 5/4 + (3^(1/2)*i)/4