Question Number 101489 by bramlex last updated on 03/Jul/20

Answered by john santu last updated on 03/Jul/20

let ((x−3))^(1/(3 )) = h and ((x+4))^(1/(3 )) = t ⇔ 5ht + 4h^2 + t^2 = 0 (4h+t)(h+t) = 0 { ((h = −t)),((4h = −t)) :} case(1) ((x−3))^(1/(3 )) = −((x+4))^(1/(3 )) ⇛ x−3 = −x−4; x = −(1/2) case(2) ((x+4))^(1/(3 )) = −4 ((x−3))^(1/(3 )) x+4 = −64(x−3) 65x = 192−4 = 188 ; x = ((188)/(65))