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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.
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Posted 10 years ago
 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 10 years ago
 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] 
Anonymous User
Anonymous User
Posted 10 years ago
 Why did you expand it?
Posted 10 years ago
 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 
Anonymous User
Anonymous User
Posted 10 years ago
 So how would I simplify and collect
Posted 10 years ago
 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.
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?