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anonymous
 one year ago
Historical Feature:
Show that the general cubic equation y^3+by^2+cy+d=0 can be transformed into an equation of the form x^3+px +q =0 by using the substitution y=xb/3
anonymous
 one year ago
Historical Feature: Show that the general cubic equation y^3+by^2+cy+d=0 can be transformed into an equation of the form x^3+px +q =0 by using the substitution y=xb/3

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\left(x  \dfrac{b}{3} \right)^3 + b\left(x  \dfrac{b}{3} \right)^2 + c\left(x  \dfrac{b}{3} \right) + d=0\) \(x^3  bx^2 + \dfrac{b^2x}{3}  \dfrac{b^3x}{27} + bx^2  \dfrac{2b^2x}{3} + \dfrac{b^3}{9} + cx  \dfrac{bc}{3} + d = 0\) \(x^3 + \dfrac{b^2x}{3}  \dfrac{b^3x}{27}  \dfrac{2b^2x}{3} + \dfrac{b^3}{9} + cx  \dfrac{bc}{3} + d=0\) \(x^3 + \left(\dfrac{b^2}{3}  \dfrac{b^3}{27}  \dfrac{2b^2}{3} + c\right)x + \dfrac{b^3}{9}  \dfrac{bc}{3} + d=0\) Where \(p = \dfrac{b^2}{3}  \dfrac{b^3}{27}  \dfrac{2b^2}{3} + c\) and \(q = \dfrac{b^3}{9}  \dfrac{bc}{3} + d\)
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