Solve #-u^3+pu-(ru)/(p+q/u-u^2)=q# For #u# ?

#-u^3+pu-(ru)/(p+q/u-u^2)=q#

1 Answer
Mar 1, 2018

#"Multiply both sides by "p+q/u-u^2" to get rid of the"#
#"denominator :"#

#u(p - u^2)(p+q/u-u^2) - r u = q(p+q/u-u^2)#

#"Multiply by "u" to have all powers positive : "#

#u(p - u^2)(p u + q - u^3) - r u^2 = q(p u + q - u^3)#

#u^6 - 2 p u^4 - q u^3 + p^2 u^2 + p q u - r u^2 = p q u + q^2 - q u^3#

#=> u^6 - 2 p u^4 + (p^2 - r) u^2 - q^2 = 0#

#"Substitute "x = u^2" to obtain a cubic equation : "#

#=> x^3 - 2 p x^2 + (p^2 - r) x - q^2 = 0#

#"If we put"#

#a = -2 p#
#b = p^2 - r#
#c = - q^2#

#"Then we have the general shape of a cubic equation : "#

#x^3 + a x^2 + b x + c = 0#

#"All that we know is that "c < 0"."#
#"So we must solve this cubic equation with a general method"#
#"like Cardano, or the substitution of Vieta, once we have the"#
#"parameters p, q, r, and hence a, b, c. We cannot give a general"#
#"formula for u, unless we give the general formula for the cubic"#
#"equation, which is very complicated."#