Is it possible to reach this expression using Mathematica's algebraic functions?

Question

I know that this expression:

-(m-2 r+b^2 (-7 m+6 r)) ((-m+r)^2+b^4 (-2 m+3 r)^2+b^2 (-3 m^2+10 m r-6 r^2))+(1+b^2)^2 m^3 Cos[t]^2

Is equal to this one:

(1-3 b^2)^3 (-m+r)^2 (-m+2 r)+(1+b^2)^2 m^3 (b^2+Cos[t]^2)

I couldn't get Mathematica to reach this simpler later form. FullSimplify and other functions combinations like FullSimplify with PowerExpand didn't work. I even used VOISImplify, and no success. Any thoughts?

Why do you think the latter one is "simpler", because it is shorter? — vapor, Mar 14 '15 at 14:49

score 3 · Accepted Answer · answered Mar 14 '15 at 15:59

3

expr = -(m - 2 r + b^2 (-7 m + 6 r)) ((-m + r)^2 + 
      b^4 (-2 m + 3 r)^2 + 
      b^2 (-3 m^2 + 10 m r - 6 r^2)) + (1 + b^2)^2 m^3 Cos[t]^2;

expr2 = (Collect[expr1 /. Cos[t]^2 -> x - b^2, x] // FullSimplify) /. 
  x -> (b^2 + Cos[t]^2)

(-1 + 3 b^2)^3 (m - 2 r) (m - r)^2 + (1 + b^2)^2 m^3 (b^2 + Cos[t]^2)

expr == expr2 // Simplify

True

LeafCount /@ {expr, expr2}

{76, 42}

answered Mar 14 '15 at 15:59

Bob Hanlon

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How did you end with that substitution? – Giovanni F. Mar 14 '15 at 16:46
2

@Giovanni - (b^2 + Cos[t]^2) was the "odd" factor in the last term of the target result so I made the expression a polynomial of that factor. – Bob Hanlon Mar 14 '15 at 16:53
Any general procedure for these cases? – Giovanni F. Mar 14 '15 at 20:32
@Giovanni - Not that I know of. – Bob Hanlon Mar 14 '15 at 21:54

Is it possible to reach this expression using Mathematica's algebraic functions?

1 Answers1