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Question Number 200886 by essaad last updated on 26/Nov/23

resoudre dans R : ((3+x))^(1/3) −((3−x))^(1/3) =((9−x^2 ))^(1/3)

resoudredansR:3+x3−3−x3=9−x23

Answered by Frix last updated on 26/Nov/23

a^(1/3) −b^(1/3) =c^(1/3) a−b−3a^(1/3) b^(1/3) (a^(1/3) −b^(1/3) )=c a−b−3a^(1/3) b^(1/3) c^(1/3) =c 27abc=(a−b−c)^3 c=ab 27a^2 b^2 =(a−b−ab)^3 a=3+x∧b=3−x 27(x−3)^2 (x+3)^2 =(x^2 +2x−9)^3 x^6 +6x^5 −42x^4 −100x^3 +621x^2 +486x−2916=0 No exact solution possible x≈−8.70670378^★ ∨x≈2.73590400 ^★ only if we use ((−r))^(1/3) =−(r)^(1/3)

a13−b13=c13a−b−3a13b13(a13−b13)=ca−b−3a13b13c13=c27abc=(a−b−c)3c=ab27a2b2=(a−b−ab)3a=3+x∧b=3−x27(x−3)2(x+3)2=(x2+2x−9)3x6+6x5−42x4−100x3+621x2+486x−2916=0Noexactsolutionpossiblex≈−8.70670378★∨x≈2.73590400★onlyifweuse−r3=−r3

Commented by essaad last updated on 26/Nov/23

thanks sir

thankssir

Answered by Rasheed.Sindhi last updated on 26/Nov/23

Another way ((3+x))^(1/3) −((3−x))^(1/3) −((9−x^2 ))^(1/3) =0 a=((3+x))^(1/3) ,b= ((3−x))^(1/3) , c=((9−x^2 ))^(1/3) a−b−c=0 determinant (((a^3 −b^3 −c^3 −3abc=(a−b−c)(a^2 +b^2 +c^2 +ab−bc+ca)))) a^3 −b^3 −c^3 −3abc=0 (3+x)−(3−x)−(9−x^2 )−3(((3+x))^(1/3) )(((3−x))^(1/3) )(((9−x^2 ))^(1/3) )=0 x^2 +2x−9=3(((9−x^2 )^2 ))^(1/3) =0 (x^2 +2x−9)^3 =27(x^2 −9)^2 x^6 +6x^5 −42x^4 −100x^3 +621x^2 +486x−2916=0 x≈−8.70670377676 , 2.73590400403

Anotherway3+x3−3−x3−9−x23=0a=3+x3,b=3−x3,c=9−x23a−b−c=0a3−b3−c3−3abc=(a−b−c)(a2+b2+c2+ab−bc+ca)a3−b3−c3−3abc=0(3+x)−(3−x)−(9−x2)−3(3+x3)(3−x3)(9−x23)=0x2+2x−9=3(9−x2)23=0(x2+2x−9)3=27(x2−9)2x6+6x5−42x4−100x3+621x2+486x−2916=0x≈−8.70670377676,2.73590400403

Answered by mr W last updated on 26/Nov/23

((3+x))^(1/3) −((3−x))^(1/3) =(((3+x)(3−x)))^(1/3) say u=((3+x))^(1/3) ,v=((3−x))^(1/3) u−v=uv ⇒v=(u/(1+u)) u^3 +v^3 =6 u^3 +(u^3 /((1+u)^3 ))=6 u^6 +3u^5 +3u^4 −4u^3 −18u^2 −18u−6=0 ⇒u≈−1.787016, 1.790059 ⇒x=u^3 −3≈−8.706704, ≈2.735906

3+x3−3−x3=(3+x)(3−x)3sayu=3+x3,v=3−x3u−v=uv⇒v=u1+uu3+v3=6u3+u3(1+u)3=6u6+3u5+3u4−4u3−18u2−18u−6=0⇒u≈−1.787016,1.790059⇒x=u3−3≈−8.706704,≈2.735906

Answered by Rasheed.Sindhi last updated on 26/Nov/23

((3+x))^(1/3) −((3−x))^(1/3) =((9−x^2 ))^(1/3) Dividing by ((9−x^2 ))^(1/3) : (1/( ((3−x))^(1/3) ))−(1/( ((3+x))^(1/3) ))=1 a=(1/( ((3−x))^(1/3) )) , b=(1/( ((3+x))^(1/3) )) ⇒ a−b−1=0 determinant (((a^3 −b^3 −1−3ab=(a−b−1)_(⇒a^3 −b^3 −1−3ab=0 ) (a^2 +b^2 +1+ab−b+a)))) (1/( 3−x ))−(1/(3+x))−1−3((1/( ((3−x))^(1/3) )))((1/( ((3+x))^(1/3) )))=0 (1/( 3−x ))−(1/(3+x))−1=3((1/( ((3−x))^(1/3) )))((1/( ((3+x))^(1/3) ))) ((3+x−3+x−9+x^2 )/(9−x^2 ))=(3/( ((9−x^2 ))^(1/3) )) x^2 +2x−9=3(9−x^2 )^(2/3) (x^2 +2x−9)^3 =27(9−x^2 )^2 x^6 +6x^5 −42x^4 −100x^3 +621x^2 +486x−2916=0 x≈−8.70670377676 , 2.73590400403

3+x3−3−x3=9−x23Dividingby9−x23:13−x3−13+x3=1a=13−x3,b=13+x3⇒a−b−1=0a3−b3−1−3ab=(a−b−1)⇒a3−b3−1−3ab=0(a2+b2+1+ab−b+a)13−x−13+x−1−3(13−x3)(13+x3)=013−x−13+x−1=3(13−x3)(13+x3)3+x−3+x−9+x29−x2=39−x23x2+2x−9=3(9−x2)2/3(x2+2x−9)3=27(9−x2)2x6+6x5−42x4−100x3+621x2+486x−2916=0x≈−8.70670377676,2.73590400403

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