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Question Number 181079 by Agnibhoo98 last updated on 21/Nov/22
Solve for x : (((x + a)/(x + b)))^3 = ((x + 2a − b)/(x − a + 2b))
Solveforx:(x+ax+b)3=x+2a−bx−a+2b
Commented by CElcedricjunior last updated on 21/Nov/22
(((x+a)/(x+b)))^3 =((x+2a−b)/(x−a+2b)) =>((x^3 +3ax^2 +3a^2 x+a^3 )/(x^3 +3b^2 x+3bx^2 +b^3 ))=((x+2a−b)/(x−a+2b)) =>x^2 (−3a^2 +6ab)+x(−3a^3 +6a^2 b)−a^4 +2a^3 b=x^2 (6b^2 a−3b^3 )+x(6ba−3b^2 )+2ab^3 −b^4 =>x^2 (−3a^2 +6ab−6b^2 a+3b^3 )+x(−3a^3 +6a^2 b−6ab+3b^2 )−a^4 +2a^3 b+2ab^3 +b^4 =0 x=((3a^2 (a−2b)+b(−3b+6a)∓(√((−3a^3 +6a^2 b−6ab+3b^2 )+4(a^4 −2a^3 b−2ab^3 −b^4 )(−3a^2 +6ab−6b^2 a+3b^3 ))))/(2(−3a^2 +6ab−6ab^2 +3b^3 ))) ...........le celebre cedric junior.....
(\boldsymbolx+\boldsymbola\boldsymbolx+\boldsymbolb)3=\boldsymbolx+2\boldsymbola−\boldsymbolb\boldsymbolx−\boldsymbola+2\boldsymbolb=>\boldsymbolx3+3\boldsymbolax2+3\boldsymbola2\boldsymbolx+\boldsymbola3\boldsymbolx3+3\boldsymbolb2\boldsymbolx+3\boldsymbolbx2+\boldsymbolb3=\boldsymbolx+2\boldsymbola−\boldsymbolb\boldsymbolx−\boldsymbola+2\boldsymbolb=>\boldsymbolx2(−3\boldsymbola2+6\boldsymbolab)+\boldsymbolx(−3\boldsymbola3+6\boldsymbola2\boldsymbolb)−\boldsymbola4+2\boldsymbola3\boldsymbolb=\boldsymbolx2(6\boldsymbolb2\boldsymbola−3\boldsymbolb3)+\boldsymbolx(6\boldsymbolba−3\boldsymbolb2)+2\boldsymbolab3−\boldsymbolb4=>\boldsymbolx2(−3\boldsymbola2+6\boldsymbolab−6\boldsymbolb2\boldsymbola+3\boldsymbolb3)+\boldsymbolx(−3\boldsymbola3+6\boldsymbola2\boldsymbolb−6\boldsymbolab+3\boldsymbolb2)−\boldsymbola4+2\boldsymbola3\boldsymbolb+2\boldsymbolab3+\boldsymbolb4=0\boldsymbolx=3\boldsymbola2(\boldsymbola−2\boldsymbolb)+\boldsymbolb(−3\boldsymbolb+6\boldsymbola)∓(−3\boldsymbola3+6\boldsymbola2\boldsymbolb−6\boldsymbolab+3\boldsymbolb2)+4(\boldsymbola4−2\boldsymbola3\boldsymbolb−2\boldsymbolab3−\boldsymbolb4)(−3\boldsymbola2+6\boldsymbolab−6\boldsymbolb2\boldsymbola+3\boldsymbolb3)2(−3\boldsymbola2+6\boldsymbolab−6\boldsymbolab2+3\boldsymbolb3)………..lecelebrecedricjunior…..
Commented by Frix last updated on 22/Nov/22
wrong.
wrong.
Answered by Rasheed.Sindhi last updated on 21/Nov/22
(((x+a)^3 )/((x+b)^3 ))=((x+2a−b)/(x−a+2b)) (p^3 /q^3 )=(u/v) ⇒((p^3 −q^3 )/(p^3 +q^3 ))=((u−v)/(u+v)) ⇒(((p−q)(p^2 +pq+q^2 ))/((p+q)(p^2 −pq+q^2 )))=((u−v)/(u+v)) p−q=x+a−x−b=a−b p+q=x+a+x+b=2x+a+b p^2 +pq+q^2 =(x^2 +2xa+a^2 )+(x^2 +(a+b)x+ab)+(x^2 +2xb+b^2 ) =3x^2 +3x(a+b)+a^2 +ab+b^2 p^2 −pq+q^2 =(x^2 +2xa+a^2 )−(x^2 +(a+b)x+ab)+(x^2 +2xb+b^2 ) =x^2 +x(a+b)+a^2 −ab+b^2 u−v=x+2a−b−x+a−2b=3(a−b) u+v=x+2a−b+x−a+2b=2x+a+b (((a−b)(3x^2 +x(a+b)+a^2 +ab+b^2 ))/((2x+a+b)(x^2 +x(a+b)+a^2 −ab+b^2 )))=((3(a−b))/(2x+a+b)) (((a−b)(3x^2 +3x(a+b)+a^2 +ab+b^2 ))/((2x+a+b)(3x^2 +x(a+b)+a^2 +ab+b^2 )))−((3(a−b))/(2x+a+b))=0 ((a−b)/(2x+a+b))(((3x^2 +3x(a+b)+a^2 +ab+b^2 )/(3x^2 +x(a+b)+a^2 +ab+b^2 ))−3)=0 ((a−b)/(2x+a+b^★ ))(((−6x^2 −2a^2 −2ab−2b^2 )/(x^2 +x(a+b)+a^2 −ab+b^(2★★) )))=0 a−b=0 ∣ −6x^2 −2a^2 −2ab−2b^2 =0 a=b ∣ 3x^2 =−(a^2 +ab+b^2 ) a=b ∣ x=±(√((−(a^2 +ab+b^2 ))/3)) ^★ 2x+a+b≠0⇒x≠−((a+b)/2) ^(★★) 3x^2 +x(a+b)+a^2 +ab+b^2 ≠0 x≠((−(a+b)±(√(a^2 +2ab+b^2 −12(a^2 +ab+b^2 ))))/6) ≠((−(a+b)±(√(−11a^2 −10ab−11b^2 )))/6)
(x+a)3(x+b)3=x+2a−bx−a+2bp3q3=uv⇒p3−q3p3+q3=u−vu+v⇒(p−q)(p2+pq+q2)(p+q)(p2−pq+q2)=u−vu+vp−q=x+a−x−b=a−bp+q=x+a+x+b=2x+a+bp2+pq+q2=(x2+2xa+a2)+(x2+(a+b)x+ab)+(x2+2xb+b2)=3x2+3x(a+b)+a2+ab+b2p2−pq+q2=(x2+2xa+a2)−(x2+(a+b)x+ab)+(x2+2xb+b2)=x2+x(a+b)+a2−ab+b2u−v=x+2a−b−x+a−2b=3(a−b)u+v=x+2a−b+x−a+2b=2x+a+b(a−b)(3x2+x(a+b)+a2+ab+b2)(2x+a+b)(x2+x(a+b)+a2−ab+b2)=3(a−b)2x+a+b(a−b)(3x2+3x(a+b)+a2+ab+b2)(2x+a+b)(3x2+x(a+b)+a2+ab+b2)−3(a−b)2x+a+b=0a−b2x+a+b(3x2+3x(a+b)+a2+ab+b23x2+x(a+b)+a2+ab+b2−3)=0a−b2x+a+b★(−6x2−2a2−2ab−2b2x2+x(a+b)+a2−ab+b2★★)=0a−b=0∣−6x2−2a2−2ab−2b2=0a=b∣3x2=−(a2+ab+b2)a=b∣x=±−(a2+ab+b2)3★2x+a+b≠0⇒x≠−a+b2★★3x2+x(a+b)+a2+ab+b2≠0x≠−(a+b)±a2+2ab+b2−12(a2+ab+b2)6≠−(a+b)±−11a2−10ab−11b26
Answered by MJS_new last updated on 21/Nov/22
x=−((a+b)/2)
x=−a+b2
Answered by mr W last updated on 21/Nov/22
if a=b: x∈R and x≠−a if a≠b: (((x+a)^3 )/((x+b)^3 ))=((x+a+(a−b))/(x+b−(a−b))) (x+a)^3 (x+b)−(a−b)(x+a)^3 =(x+b)^3 (x+a)+(a−b)(x+b)^3 (x+a)(x+b)[(x+a)^2 −(x+b)^2 ]=(a−b)(x+b)^3 +(a−b)(x+a)^3 (x+a)(x+b)[(x+a)+(x+b)]=(x+b)^3 +(x+a)^3 let u=x+a, v=x+b uv(u+v)=u^3 +v^3 =(u+v)(u^2 −uv+v^2 ) (u+v)(u−v)^2 =0 (2x+a+b)(a−b)^2 =0 ⇒2x+a+b=0 ⇒x=−((a+b)/2)
ifa=b:x∈Randx≠−aifa≠b:(x+a)3(x+b)3=x+a+(a−b)x+b−(a−b)(x+a)3(x+b)−(a−b)(x+a)3=(x+b)3(x+a)+(a−b)(x+b)3(x+a)(x+b)[(x+a)2−(x+b)2]=(a−b)(x+b)3+(a−b)(x+a)3(x+a)(x+b)[(x+a)+(x+b)]=(x+b)3+(x+a)3letu=x+a,v=x+buv(u+v)=u3+v3=(u+v)(u2−uv+v2)(u+v)(u−v)2=0(2x+a+b)(a−b)2=0⇒2x+a+b=0⇒x=−a+b2
Answered by Frix last updated on 22/Nov/22
x≠−b∧x≠a−2b (x+a)^3 (x−a+2b)−(x+b)^3 (x+2a−b)=0 (−2a^3 +6a^2 b−6ab^2 +2b^3 )x−a^4 +2a^3 b−2ab^3 +b^4 =0 −2(a−b)^3 x−(a+b)(a−b)^3 =0 (a−b)^3 (2x+a+b)=0 ⇒ (a=b∨x=−((a+b)/2))∧x≠−b∧x≠a−2b ⇔ (a=b∧x∈R\{−b, a−2b})∨(a≠b∧x=−((a+b)/2))
x≠−b∧x≠a−2b(x+a)3(x−a+2b)−(x+b)3(x+2a−b)=0(−2a3+6a2b−6ab2+2b3)x−a4+2a3b−2ab3+b4=0−2(a−b)3x−(a+b)(a−b)3=0(a−b)3(2x+a+b)=0⇒(a=b∨x=−a+b2)∧x≠−b∧x≠a−2b⇔(a=b∧x∈R∖{−b,a−2b})∨(a≠b∧x=−a+b2)
Answered by manxsol last updated on 22/Nov/22
(((x+a)/(x+b)))^3 =((x+a+a−b)/( x+b−(a−b))) (((x+a)/(x+b)))^3 =(((x+a)+[(x+a)−(x+b)])/((x+b)−[(x+a)−(x−b)])) x+a=p x+b=q (p^3 /q^3 )=((2p−q)/(2q−p)) 2qp^3 −p^4 =2pq^3 −q^4 2qp(p^2 −q^2 )=p^4 −q^4 (p^2 −q^2 )(p^2 −2qp+q^2 )=0 (p−q)(p+q)(p−q)^2 =0 (p−q)^3 (p+q)=0 p=q⇒x+a=x+b⇒a=b p=−q⇒x+a^ =−x−b x=−(((a+b))/2) resumen determinant (((restriction),(solution)),((x≠{−b;a−2b}),(x=−((a+b)/2))),((a=b;x≠−b),(ℜ−{−b})))
(x+ax+b)3=x+a+a−bx+b−(a−b)(x+ax+b)3=(x+a)+[(x+a)−(x+b)](x+b)−[(x+a)−(x−b)]x+a=px+b=qp3q3=2p−q2q−p2qp3−p4=2pq3−q42qp(p2−q2)=p4−q4(p2−q2)(p2−2qp+q2)=0(p−q)(p+q)(p−q)2=0(p−q)3(p+q)=0p=q⇒x+a=x+b⇒a=bp=−q⇒x+a=−x−bx=−(a+b)2resumenrestrictionsolutionx≠{−b;a−2b}x=−a+b2a=b;x≠−bℜ−{−b}
Answered by Rasheed.Sindhi last updated on 22/Nov/22
(((x + a)/(x + b)))^3 = ((x + 2a − b)/(x − a + 2b)) ; x=? determinant (((A=((A+B)/2)+((A−B)/2) _(B=((A+B)/2)−((A−B)/2)^( ) ) ^( ) ))) •a=((a+b)/2)+((a−b)/2) •b=((a+b)/2)−((a−b)/2) •2a−b=(((2a−b)+(−a+2b))/2)+(((2a−b)−(−a+2b))/2) =((a+b)/2)+3(((a−b)/2)) •−a+2b=(((2a−b)+(−a+2b))/2)−(((2a−b)−(−a+2b))/2) =((a+b)/2)−3(((a−b)/2)) (((x+((a+b)/2)+((a−b)/2))^3 )/((x+((a+b)/2)−((a−b)/2))^3 ))=((x+((a+b)/2)+3(((a−b)/2)))/(x+((a+b)/2)−3(((a−b)/2)))) x+((a+b)/2)=y , ((a−b)/2)=k (((y+k)^3 )/((y−k)^3 ))=((y+3k)/(y−3k)) (((y+k)^2 )/((y−k)^2 ))=(((y−k)(y+3k))/((y+k)(y−3k))) ((y^2 +2ky+k^2 )/(y^2 −2ky+k^2 ))−1=((y^2 +2ky−3k^2 )/(y^2 −2ky−3k^2 ))−1 ((4ky)/(y^2 −2ky+k^2 ))=((4ky)/(y^2 −2ky−3k^2 )) (y/(y^2 −2ky+k^2 ))−(y/(y^2 −2ky−3k^2 ))=0 ;k≠0 y((1/(y^2 −2ky+k^2 ))−(1/(y^2 −2ky−3k^2 )))=0 y=0^★ ∣ (1/(y^2 −2ky+k^2 ))−(1/(y^2 −2ky−3k^2 ))=0^⧫ ^★ y=0⇒x+((a+b)/2)=0⇒x=−((a+b)/2) ^⧫ (1/(y^2 −2ky+k^2 ))−(1/(y^2 −2ky−3k^2 ))=0 (1/(y^2 −2ky+k^2 ))=(1/(y^2 −2ky−3k^2 )) y^2 ^(×) −2ky^(×) −3k^2 =y^2 ^(×) −2ky^(×) +k^2 4k^2 =0⇒k=0⇒((a−b)/2)=0⇒a=b
(x+ax+b)3=x+2a−bx−a+2b;x=?A=A+B2+A−B2B=A+B2−A−B2∙a=a+b2+a−b2∙b=a+b2−a−b2∙2a−b=(2a−b)+(−a+2b)2+(2a−b)−(−a+2b)2=a+b2+3(a−b2)∙−a+2b=(2a−b)+(−a+2b)2−(2a−b)−(−a+2b)2=a+b2−3(a−b2)(x+a+b2+a−b2)3(x+a+b2−a−b2)3=x+a+b2+3(a−b2)x+a+b2−3(a−b2)x+a+b2=y,a−b2=k(y+k)3(y−k)3=y+3ky−3k(y+k)2(y−k)2=(y−k)(y+3k)(y+k)(y−3k)y2+2ky+k2y2−2ky+k2−1=y2+2ky−3k2y2−2ky−3k2−14kyy2−2ky+k2=4kyy2−2ky−3k2yy2−2ky+k2−yy2−2ky−3k2=0;k≠0y(1y2−2ky+k2−1y2−2ky−3k2)=0y=0★∣1y2−2ky+k2−1y2−2ky−3k2=0⧫★y=0⇒x+a+b2=0⇒x=−a+b2⧫1y2−2ky+k2−1y2−2ky−3k2=01y2−2ky+k2=1y2−2ky−3k2y2×−2ky×−3k2=y2×−2ky×+k24k2=0⇒k=0⇒a−b2=0⇒a=b

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