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x-4-2-1-3-4-x-3-2-1-3-5-x-2-x-12-1-3-0-

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Question Number 96131 by bemath last updated on 30/May/20
(((x+4)^2 ))^(1/(3 )) + 4 (((x−3)^2 ))^(1/(3 )) + 5 ((x^2 +x−12))^(1/(3 )) = 0
(x+4)23+4(x−3)23+5x2+x−123=0
Answered by john santu last updated on 30/May/20
let ((x+4))^(1/(3 )) = u & ((x−3))^(1/(3 )) = v ⇒u^2 +4v^2 +5uv = 0 (u+4v)(u+v) = 0 { ((u = −v)),((u=−4v)) :} (1) ((x+4))^(1/(3 )) = −((x−3))^(1/(3 )) x+4 = −x+3 ⇒ x=−(1/2) (2) ((x+4))^(1/(3 )) = −4 ((x−3))^(1/(3 )) x+4 = −64x+192 65x = 184 ; x = ((184)/(65)) .
letx+43=u&x−33=v⇒u2+4v2+5uv=0(u+4v)(u+v)=0{u=−vu=−4v(1)x+43=−x−33x+4=−x+3⇒x=−12(2)x+43=−4x−33x+4=−64x+19265x=184;x=18465.

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