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.
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]
Why did you expand it?
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
So how would I simplify and collect
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.
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?
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]