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In the following simplification why doesn't Mathematica get a zero when asked to compute "Y-Z" eventually? 

 a = 1 - 4 A Q^2; 
 b = (-972 + 648) A Q^2 + 54;
 c = 9 - 36 A Q^2; 

 Y = 12 Q/( 
 Sqrt[A] (  
   2 - (3 2^(1/3) a )/(Sqrt[b^2 - 4 c^3] + b  )^(1/3)  - ( 
       Sqrt[b^2 - 4 c^3] + b  )^(1/3)/( 3 2^(1/3) )     )   );

 Z = (4/A) + (3 2^(1/3) a )/(A (Sqrt[b^2 - 4 c^3] + b  )^(1/3) )  +  ( 
  Sqrt[b^2 - 4 c^3] + b  )^(1/3)/( 3 2^(1/3) A);

Q = Sqrt [ 3/(16 A)];
Y - Z // FullSimplify
(-9 + 4 Sqrt[3] Sqrt[1/A] Sqrt[A])/(2 A)

One can do the above calculation by hand and one would see that for the specific chosen value of $Q = \sqrt{3/16 A}$ one would get $Y = Z = \frac{3}{A}$. Why doesn't Mathematica see this?

  • And is there a way to get Mathematica to detect if there are other values of $Q$ where $Y = Z$? (..I found this one value special value of $Q$ by just staring at the equation for sometime...)

Rewriting the functions again.

$Y = \frac{12Q}{\sqrt{A}\sqrt{ 2 - \frac{3 (2^{1/3})a }{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} } - \frac{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} }{3 (2^{1/3})} } }$

$Z = \frac{1}{A} \left (4 + \frac{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} }{3 (2^{1/3})} + \frac{3 (2^{1/3})a }{(\sqrt{b^2 - 4 c^3 } + b )^{1/3} }\right ) $

at the chosen values of $Q = \sqrt{\frac{3}{16A} }$ one has $a = \frac{1}{4}$ and $b = - \frac{27}{4}$ and $b^2 - 4 c^3 = 0$. Also whenever I encounter $b^{1/3}$ I am writing that as $-3 \times 2^{-2/3}$. Substituting these into the above one gets $Y = Z = \frac{3}{A}$

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  • $\begingroup$ You can try $Assumptions = {A > 0}; Y - Z // FullSimplify but it seems there is a mistake somewhere. $\endgroup$
    – Kuba
    Commented Jun 26, 2014 at 10:28
  • $\begingroup$ Getting $\frac{4 \sqrt{3} \sqrt{\frac{1}{A}} \sqrt{A}-9}{2 A}$ and with assuming A > 0 $\frac{4 \sqrt{3}-9}{2 A}$. No Zero. $\endgroup$
    – Phab
    Commented Jun 26, 2014 at 10:36
  • $\begingroup$ @Phab But did you do the calculation by hand? Isn't it zero doing so? $\endgroup$
    – Student
    Commented Jun 26, 2014 at 11:35
  • $\begingroup$ @Anirbit Not on the first try. No time for a second so far. $\endgroup$
    – Phab
    Commented Jun 26, 2014 at 12:36
  • $\begingroup$ @Phab I am quite confident that it is zero. (there are other cross-checks which it passes given the larger context fro where it comes) It would be great if you could kindly check again. $\endgroup$
    – Student
    Commented Jun 26, 2014 at 13:09

1 Answer 1

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The problem is that b^(1/3) has three roots. You are choosing the real root, -(3/2^(2/3)), while Mathematica is choosing, (3 (-1)^(1/3))/2^(2/3) which is complex. To get the result you want, rewrite your expressions for Y and Z like this:

Y =
  12 Q/
    Sqrt[A (2 - 
            (3 2^(1/3) a)/Surd[Sqrt[b^2 - 4 c^3] + b, 3] - 
            Surd[Sqrt[b^2 - 4 c^3] + b, 3]/(3 2^(1/3)))];

Z = (4/A) + (3 2^(1/3) a)/A /Surd[Sqrt[b^2 - 4 c^3] + b, 3] + 
   Surd[Sqrt[b^2 - 4 c^3] + b, 3]/(3 2^(1/3) A);

With the above you will get

FullSimplify[Y - Z, Assumptions -> A > 0]

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