Question Number 100561 by DGmichael last updated on 27/Jun/20

Commented bybramlex last updated on 27/Jun/20

x−y = ((x^3 −y^3 )/(x^2 +xy+y^2 )) x=((3+(√(9+((125)/(27)))) ))^(1/(3 )) , y = ((−3+(√(9+((125)/(27))))))^(1/(3 )) x−y = (6/((((3+(√(9+((125)/(27)))))^2 ))^(1/(3 )) +(((125)/(27)))^(1/(3 )) +(((−3+(√(9+((125)/(27)))))^2 ))^(1/(3 )) )) = (6/((5/3)+((18+((125)/(27)) + 6(√(9+((125)/(27))))))^(1/(3 )) +((18+((125)/(27))−6(√(9+((125)/(27))))))^(1/(3 )) ))

Answered by MJS last updated on 27/Jun/20

A is the solution of A^3 +pA+q=0 Cardano A=((−(q/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/3) +((−(q/2)−(√((q^2 /4)+(p^3 /(27))))))^(1/3) = =((−(q/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/3) −(((q/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/3) (q/2)=3 ⇒ q=6 (q^2 /4)=9 ⇒ q=±6 ⇒ q=6 (p^3 /(27))=((125)/(27)) ⇒ p=5 A^3 +5A+6=0 ⇒ A_1 =−1∧A_(2, 3) =(1/2)±((√(23))/2)i ⇒ A=A_1 =−1