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I have an expression which reads

f1[kx_, ky_, kz_] = \[Nu]^3*m1 // Simplify;
m1= (2 d (kx kz + I ky Sqrt[kx^2 + ky^2 + kz^2])^2 (d (-I kx^10 (ky^2 + kz^2) + 
    I ky^6 kz^2 (ky^2 + kz^2)^2 - 
    4 kx^7 ky kz^3 Sqrt[kx^2 + ky^2 + kz^2] + 
    kx ky^5 kz Sqrt[
     kx^2 + ky^2 + kz^2] (ky^4 - 4 ky^2 kz^2 - 5 kz^4) + 
    kx^5 ky kz Sqrt[kx^2 + ky^2 + kz^2] (ky^4 - kz^4) - 
    I kx^8 (3 ky^4 + kz^4) + 
    2 kx^3 ky^3 kz Sqrt[kx^2 + ky^2 + kz^2] (ky^4 + 5 kz^4) - 
    I kx^6 (3 ky^6 - 4 ky^4 kz^2 - 7 ky^2 kz^4) - 
    I kx^4 (ky^8 - 5 ky^6 kz^2 + ky^4 kz^4 - 5 ky^2 kz^6) + 
    I kx^2 (3 ky^8 kz^2 - 7 ky^6 kz^4 - 10 ky^4 kz^6)) + 
    Sqrt[3] kx (2 kx^8 ky kz + 3 kx^6 ky kz (ky^2 + kz^2) + 
    ky^5 kz (ky^2 + kz^2)^2 - 
    I kx^7 (ky^2 - kz^2) Sqrt[kx^2 + ky^2 + kz^2] - 
    2 I kx^5 ky^2 (ky^2 + kz^2) Sqrt[kx^2 + ky^2 + kz^2] + 
    4 I kx ky^4 kz^2 (ky^2 + kz^2) Sqrt[kx^2 + ky^2 + kz^2] + 
    kx^2 ky^3 kz (ky^4 - 5 ky^2 kz^2 - 6 kz^4) + 
    kx^4 ky kz (ky^4 - 4 ky^2 kz^2 + kz^4) - 
    I kx^3 ky^2 Sqrt[
     kx^2 + ky^2 + 
      kz^2] (ky^4 - ky^2 kz^2 + 4 kz^4)) \[Nu]))/((kx^2 + 
 ky^2)^2 (kx^2 + ky^2 + kz^2)^(3/2) (I kx kz + 
 ky Sqrt[kx^2 + ky^2 + kz^2])^2 \[Nu] (-2 Sqrt[3]
   d ky kz (-2 kx^2 + ky^2 + kz^2) + 3 I kx (ky^2 + kz^2) \[Nu]))

Here $\nu$ and $d$ are constant variables. I would like to evaluate this expression at $\nu \sqrt{k_x^2 + k_y^2 +k_z^2}=\omega$, where $\omega$ is a constant. I have tried Eliminate and Reduce in this thread but they are very slow and I was not bale to evaluate my expression using them. Do you have a any suggestion?

Edit:

I have applied the suggestion in the comments to the denominator of my expression as

 smp = Simplify[Denominator[f1[kx, ky, kz]], \[Nu]*Sqrt[kx^2 + ky^2 + kz^2] == 1];
 smp - Denominator[f1[kx, ky, kz]]
 (*0*)

The simplified expression is exactly the same as the denominator of $f_1$ even though I set $\omega=1$.

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  • $\begingroup$ I am looking for a final expression in terms of $\omega$, $d$, and $\nu$. Your suggested simplification won't remove all of the k variables. $\endgroup$
    – Shasa
    Sep 17, 2020 at 9:46
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    $\begingroup$ You only have one constraint , how can you expect to remove all k variables? $\endgroup$ Sep 17, 2020 at 9:50
  • $\begingroup$ Correct. I have to simplify my expression as much as I can. $\endgroup$
    – Shasa
    Sep 17, 2020 at 9:56
  • $\begingroup$ Thank you, Bill. This works a bit better but there are still some terms like $(k_x^2 + k_y^2 + k_z^2)^(3/2)$ which have not been replaced by $\omgea^3$. Do you know how I get them? $\endgroup$
    – Shasa
    Sep 17, 2020 at 11:38

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