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Question: How is the simplification achieved

Question:How is the simplification achieved

Ronan 1311 Maple 2024
radical simplification
December 20 2024
0

2024-12-20_Q_simplification_Question.mw
Solve the general cubic. Apply values and simplify. 

Could someone show how Maple simplifies to the value of X=3? I tried doing it manually and I could not figure it out. 

Also is there a Help assistant to see the setps?

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restart
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X^3+a*X=b

X^3+X*a = b

 (1)
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sol:=solve(X^3+a*X=b,[X])

[[X = (1/6)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)-2*a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)], [X = -(1/12)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+((1/2)*I)*3^(1/2)*((1/6)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+2*a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3))], [X = -(1/12)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)-((1/2)*I)*3^(1/2)*((1/6)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+2*a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3))]]

 (2)
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vals:=[a=6,b=45]

[a = 6, b = 45]

 (3)
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Nans:=(map(eval,sol,vals))

[[X = (1/6)*(4860+12*166617^(1/2))^(1/3)-12/(4860+12*166617^(1/2))^(1/3)], [X = -(1/12)*(4860+12*166617^(1/2))^(1/3)+6/(4860+12*166617^(1/2))^(1/3)+((1/2)*I)*3^(1/2)*((1/6)*(4860+12*166617^(1/2))^(1/3)+12/(4860+12*166617^(1/2))^(1/3))], [X = -(1/12)*(4860+12*166617^(1/2))^(1/3)+6/(4860+12*166617^(1/2))^(1/3)-((1/2)*I)*3^(1/2)*((1/6)*(4860+12*166617^(1/2))^(1/3)+12/(4860+12*166617^(1/2))^(1/3))]]

 (4)
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simplify(Nans)

[[X = 3], [X = (1/4)*(I*3^(1/2)*(180+44*17^(1/2))^(2/3)+(8*I)*3^(1/2)-(180+44*17^(1/2))^(2/3)+8)/(180+44*17^(1/2))^(1/3)], [X = -3/2-((1/2)*I)*51^(1/2)]]

 (5)
 

 

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