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How would I solve this

Anonymous User
Anonymous User
Posted 10 years ago

How would I simplify the result and collect the terms of ascending powers of x? Consider the expression (z^2 - qz - 1/27 p^3)/z. Use replacement rules to replace z by w^3, then w by 1/6 (3 y + Sqrt[3] Sqrt[4 p + 3 y^2]), and then p by (b - a^2/3) and q by ((-2 a^3 + 9 ab - 27 c)/27), and finally, y by (x + a/3). Then simplify the result and collect the terms as ascending powers of x.

POSTED BY: Anonymous User
7 Replies
Collect[ExpandAll[
  Simplify[((((((z^2 - qz - 1/27 p^3)/z) /. z -> w^3) /. 
          w -> 1/6 (3 y + Sqrt[3] Sqrt[4 p + 3 y^2])) /. 
        p -> (b - a^2/3)) /. q -> ((-2 a^3 + 9 ab - 27 c)/27)) /. 
    y -> (x + a/3)]], x]
POSTED BY: Udo Krause
Anonymous User
Anonymous User
Posted 10 years ago

Why did you expand it?

POSTED BY: Anonymous User

You are right, Collect[] seems to be enough

In[2]:= Product[x - o, {o, 0, 9}]
Out[2]= (-9 + x) (-8 + x) (-7 + x) (-6 + x) (-5 + x) (-4 + x) (-3 + x) (-2 + x) (-1 + x) x

In[3]:= Collect[Product[x - o, {o, 0, 9}], x]
Out[3]= -362880 x + 1026576 x^2 - 1172700 x^3 + 723680 x^4 - 
         269325 x^5 + 63273 x^6 - 9450 x^7 + 870 x^8 - 45 x^9 + x^10
POSTED BY: Udo Krause
Anonymous User
Anonymous User
Posted 10 years ago

So how would I simplify and collect

POSTED BY: Anonymous User

So how would I simplify and collect

What do you mean by how? Use Mathematica as mentioned above. If you don't have Mathematica at your disposal it will be a rather long and tedious piece of substantial algebra to do - that's why Wolfram inventend Mathematica.

POSTED BY: Udo Krause
Anonymous User
Anonymous User
Posted 10 years ago

I'm sorry I am new to Mathematica. I just want to know if you excluded the ExpandAll command, what would the initial input look like?

POSTED BY: Anonymous User

Just omit the ExpandAll[] call, with other words

Collect[
  Simplify[((((((z^2 - qz - 1/27 p^3)/z) /. z -> w^3) /. 
          w -> 1/6 (3 y + Sqrt[3] Sqrt[4 p + 3 y^2])) /. 
        p -> (b - a^2/3)) /. q -> ((-2 a^3 + 9 ab - 27 c)/27)) /. 
    y -> (x + a/3)], x]

in any case you have x in the numerator as well as in the denominator. You can use ExpandDenominator as well as ExpandNominator to control the expansion in detail if needed. You can even include the Simplify in the Collect, then it reads

Collect[((((((z^2 - qz - 1/27 p^3)/z) /. z -> w^3) /. 
              w -> 1/6 (3 y + Sqrt[3] Sqrt[4 p + 3 y^2])) /. 
            p -> (b - a^2/3)) /. q -> ((-2 a^3 + 9 ab - 27 c)/27)) /. 
        y -> (x + a/3), x, Simplify]
POSTED BY: Udo Krause
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