Menu Close

a-a-b-b-b-c-c-c-a-4-ab-2-bc-2-abc-ca-2-a-2-b-b-2-c-c-2-a-a-a-b-3-b-b-c-3-c-c-a-3-

Tinku Tara 0 Comments
Pin




Question Number 60984 by naka3546 last updated on 28/May/19
(a/(a−b)) + (b/(b−c)) + (c/(c−a)) = 4 ab^2 + bc^2 + abc + ca^2 = a^2 b + b^2 c + c^2 a ((a/(a−b)))^3 + ((b/(b−c)))^3 + ((c/(c−a)))^3 = ?
aa−b+bb−c+cc−a=4ab2+bc2+abc+ca2=a2b+b2c+c2a(aa−b)3+(bb−c)3+(cc−a)3=?
Commented by naka3546 last updated on 28/May/19
73 ?
73?
Commented by Prithwish sen last updated on 28/May/19
ab^2 +bc^2 +abc+ca^2 =a^2 b+b^2 c+c^2 a ab(a−b)+bc(b−c)+ac(c−a)=abc (((a−b))/c)+(((b−c))/a)+(((c−a))/b)=1 ((−[(b−c)+(c−a)])/c)+ (((b−c))/a)+(((c−a))/b)=1 (b−c)[(1/a)−(1/c)] +(c−a)[(1/b)−(1/c)] =1 (((b−c)(c−a))/c)[(1/a)−(1/b)]=1 ((abc)/((a−b)(b−c)(c−a)))=−1 Now let A=(a/((a−b))), B=(b/((b−c))), C=(c/((c−a))) ∴ A+B+C=4, ABC=−1 we have to calculate A^3 +B^3 +C^3 please help.
ab2+bc2+abc+ca2=a2b+b2c+c2aab(a−b)+bc(b−c)+ac(c−a)=abc(a−b)c+(b−c)a+(c−a)b=1−[(b−c)+(c−a)]c+(b−c)a+(c−a)b=1(b−c)[1a−1c]+(c−a)[1b−1c]=1(b−c)(c−a)c[1a−1b]=1abc(a−b)(b−c)(c−a)=−1NowletA=a(a−b),B=b(b−c),C=c(c−a)∴A+B+C=4,ABC=−1wehavetocalculateA3+B3+C3pleasehelp.
Commented by naka3546 last updated on 28/May/19
(a/(a−b)) + (b/(b−c)) + (c/(c−a)) = 4 ⇒ 3 + ((b/(a−b)) + (c/(b−c)) + (a/(c−a))) = 4 ⇔ ((b/(a−b)) + (c/(b−c)) + (a/(c−a))) = 1 ⇒ ((abc − (a^2 b + b^2 c + c^2 a))/((ab^2 + bc^2 + ca^2 ) − (a^2 b + b^2 c + c^2 a))) = 1 ⇔ abc = ab^2 + bc^2 + ca^2 ... (i) abc + ab^2 + bc^2 + ca^2 = a^2 b + b^2 c + c^2 a ⇔ a^2 b + b^2 c + c^2 a = 2abc ... (ii) ⇔ (a−b)(b−c)(c−a) = −abc ... (iii) ((a/(a−b)))^3 + ((b/(b−c)))^3 + ((c/(c−a)))^3 = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 − 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))(((ab)/((a−b)(b−c))) + ((bc)/((b−c)(c−a))) + ((ca)/((c−a)(a−b)))) + 3(((abc)/((a−b)(b−c)(c−a)))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 − 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))(− ((c−a)/c) − ((a−b)/a) − ((b−c)/b)) + 3(((abc)/((a−b)(b−c)(c−a)))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( ((c−a)/c) + ((a−b)/a) + ((b−c)/b)) + 3(((abc)/(−abc))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( 3 − ((a/c) + (b/a) + (c/b))) + 3(((abc)/(−abc))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( 3 − (((a^2 b + b^2 c + c^2 a)/(abc)))) − 3 = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( 3 − (((2abc)/(abc)))) − 3 = (4)^3 + 3(4)( 3 − 2) − 3 = 64 + 12 − 3 = 73
aa−b+bb−c+cc−a=4⇒3+(ba−b+cb−c+ac−a)=4⇔(ba−b+cb−c+ac−a)=1⇒abc−(a2b+b2c+c2a)(ab2+bc2+ca2)−(a2b+b2c+c2a)=1⇔abc=ab2+bc2+ca2…(i)abc+ab2+bc2+ca2=a2b+b2c+c2a⇔a2b+b2c+c2a=2abc…(ii)⇔(a−b)(b−c)(c−a)=−abc…(iii)(aa−b)3+(bb−c)3+(cc−a)3=(aa−b+bb−c+cc−a)3−3(aa−b+bb−c+cc−a)(ab(a−b)(b−c)+bc(b−c)(c−a)+ca(c−a)(a−b))+3(abc(a−b)(b−c)(c−a))=(aa−b+bb−c+cc−a)3−3(aa−b+bb−c+cc−a)(−c−ac−a−ba−b−cb)+3(abc(a−b)(b−c)(c−a))=(aa−b+bb−c+cc−a)3+3(aa−b+bb−c+cc−a)(c−ac+a−ba+b−cb)+3(abc−abc)=(aa−b+bb−c+cc−a)3+3(aa−b+bb−c+cc−a)(3−(ac+ba+cb))+3(abc−abc)=(aa−b+bb−c+cc−a)3+3(aa−b+bb−c+cc−a)(3−(a2b+b2c+c2aabc))−3=(aa−b+bb−c+cc−a)3+3(aa−b+bb−c+cc−a)(3−(2abcabc))−3=(4)3+3(4)(3−2)−3=64+12−3=73
Commented by MJS last updated on 28/May/19
how do you get (i)? (b/(a−b))+(c/(b−c))+(a/(c−a))≠((abc−(a^2 b+b^2 c+c^2 a))/((ab^2 +bc^2 +ca^2 )−(a^2 b+b^2 c+c^2 a)))
howdoyouget(i)?ba−b+cb−c+ac−a≠abc−(a2b+b2c+c2a)(ab2+bc2+ca2)−(a2b+b2c+c2a)
Commented by Prithwish sen last updated on 28/May/19
great sir
greatsir
Commented by naka3546 last updated on 28/May/19
ab^2 + bc^2 + ca^2 + abc = a^2 b + b^2 c + c^2 a ⇒ (a+b+c)(ab+bc+ca) − (2abc + ab^2 + bc^2 + ca^2 ) = a^2 b + b^2 c + c^2 a ⇔ (a+b+c)(ab+bc+ca) = 2(a^2 b + b^2 c + c^2 a) (b/(a−b)) + (c/(b−c)) + (a/(c−a)) = 1 ⇒ ((b(b−c)(c−a) + c(a−b)(c−a) + a(a−b)(b−c))/((a−b)(b−c)(c−a))) = 1 ⇒ ((b^2 c − bc^2 − ab^2 + abc + c^2 a − bc^2 − ca^2 + abc + a^2 b − ab^2 − ca^2 + abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((a^2 b + b^2 c + c^2 a) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((ab^2 + bc^2 + ca^2 + abc) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((4abc) − (a^2 b + b^2 c + c^2 a))/((ab^2 + bc^2 + ca^2 ) − (a^2 b + b^2 c + c^2 a))) = 1 4abc = ab^2 + bc^2 + ca^2 ⇔ a^2 b + b^2 c + c^2 a = 5abc May be there are mistakes . Typo , sir.
ab2+bc2+ca2+abc=a2b+b2c+c2a⇒(a+b+c)(ab+bc+ca)−(2abc+ab2+bc2+ca2)=a2b+b2c+c2a⇔(a+b+c)(ab+bc+ca)=2(a2b+b2c+c2a)ba−b+cb−c+ac−a=1⇒b(b−c)(c−a)+c(a−b)(c−a)+a(a−b)(b−c)(a−b)(b−c)(c−a)=1⇒b2c−bc2−ab2+abc+c2a−bc2−ca2+abc+a2b−ab2−ca2+abc(a−b)(b−c)(c−a)=1⇒(a2b+b2c+c2a)−2(ab2+bc2+ca2)+3abc(a−b)(b−c)(c−a)=1⇒(ab2+bc2+ca2+abc)−2(ab2+bc2+ca2)+3abc(a−b)(b−c)(c−a)=1⇒(4abc)−(a2b+b2c+c2a)(ab2+bc2+ca2)−(a2b+b2c+c2a)=14abc=ab2+bc2+ca2⇔a2b+b2c+c2a=5abcMaybetherearemistakes.Typo,sir.
Commented by naka3546 last updated on 28/May/19
ab^2 + bc^2 + ca^2 + abc = a^2 b + b^2 c + c^2 a ⇒ (a+b+c)(ab+bc+ca) − (2abc + ab^2 + bc^2 + ca^2 ) = a^2 b + b^2 c + c^2 a ⇔ (a+b+c)(ab+bc+ca) = 2(a^2 b + b^2 c + c^2 a) (b/(a−b)) + (c/(b−c)) + (a/(c−a)) = 1 ⇒ ((b(b−c)(c−a) + c(a−b)(c−a) + a(a−b)(b−c))/((a−b)(b−c)(c−a))) = 1 ⇒ ((b^2 c − bc^2 − ab^2 + abc + c^2 a − bc^2 − ca^2 + abc + a^2 b − ab^2 − ca^2 + abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((a^2 b + b^2 c + c^2 a) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((ab^2 + bc^2 + ca^2 + abc) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((4abc) − (a^2 b + b^2 c + c^2 a))/((ab^2 + bc^2 + ca^2 ) − (a^2 b + b^2 c + c^2 a))) = 1 4abc = ab^2 + bc^2 + ca^2 ⇔ a^2 b + b^2 c + c^2 a = 5abc May be there are mistakes . Typo , sir.
ab2+bc2+ca2+abc=a2b+b2c+c2a⇒(a+b+c)(ab+bc+ca)−(2abc+ab2+bc2+ca2)=a2b+b2c+c2a⇔(a+b+c)(ab+bc+ca)=2(a2b+b2c+c2a)ba−b+cb−c+ac−a=1⇒b(b−c)(c−a)+c(a−b)(c−a)+a(a−b)(b−c)(a−b)(b−c)(c−a)=1⇒b2c−bc2−ab2+abc+c2a−bc2−ca2+abc+a2b−ab2−ca2+abc(a−b)(b−c)(c−a)=1⇒(a2b+b2c+c2a)−2(ab2+bc2+ca2)+3abc(a−b)(b−c)(c−a)=1⇒(ab2+bc2+ca2+abc)−2(ab2+bc2+ca2)+3abc(a−b)(b−c)(c−a)=1⇒(4abc)−(a2b+b2c+c2a)(ab2+bc2+ca2)−(a2b+b2c+c2a)=14abc=ab2+bc2+ca2⇔a2b+b2c+c2a=5abcMaybetherearemistakes.Typo,sir.
Commented by naka3546 last updated on 28/May/19
37 ?
37?
Commented by MJS last updated on 28/May/19
I′ll have to look into it again...
I′llhavetolookintoitagain…
Answered by tanmay last updated on 28/May/19
(a/(a−b))+(b/(b−c))+(c/(c−a))=4 (1/(1−(b/a)))+(1/(1−(c/b)))+(1/(1−(a/c)))=4 u=(b/a) v=(c/b) w=(a/c) uvw=1 (1/(1−u))+(1/(1−v))+(1/(1−w))=4 (1/(1−u))−1+(1/(1−v))−1+(1/(1−w))−1=1 (u/(1−u))+(v/(1−v))+(w/(1−w))=1....(1) ((1/(1−u)))^3 +((1/(1−v)))^3 +((1/(1−w)))^3 =??? ab^2 +bc^2 +ca^2 +abc−a^2 b−b^2 c−c^2 a=0 ab(b−a)+bc(c−b)+ca(a−c)+abc=0 a^2 ×b((b/a)−1)+b^2 ×c((c/b)−1)+c^2 a((a/c)−1)+abc=0 a^3 ×(b/a)((b/a)−1)+b^3 ×(c/b)((c/b)−1)+c^3 ×(a/c)((a/c)−1)+abc=0 a^2 b(u−1)+b^2 c(v−1)+c^2 a(w−1)+abc=0 wait... u=(b/a) v=(c/b) w=(a/c) uvw=1 u:v:w=(b/a):(c/b):(a/c)=b^2 c:ac^2 :a^2 b b^2 c=uk ac^2 =vk a^2 b=wk a^3 b^3 c^3 =uvwk^3 abc=(uvw)^(1/3) k abc=k wk(u−1)+uk(v−1)+vk(w−1)+k=0 uwk−wk+uvk−uk+vwk−vk+k=0
aa−b+bb−c+cc−a=411−ba+11−cb+11−ac=4u=bav=cbw=acuvw=111−u+11−v+11−w=411−u−1+11−v−1+11−w−1=1u1−u+v1−v+w1−w=1….(1)(11−u)3+(11−v)3+(11−w)3=???ab2+bc2+ca2+abc−a2b−b2c−c2a=0ab(b−a)+bc(c−b)+ca(a−c)+abc=0a2×b(ba−1)+b2×c(cb−1)+c2a(ac−1)+abc=0a3×ba(ba−1)+b3×cb(cb−1)+c3×ac(ac−1)+abc=0a2b(u−1)+b2c(v−1)+c2a(w−1)+abc=0wait…u=bav=cbw=acuvw=1u:v:w=ba:cb:ac=b2c:ac2:a2bb2c=ukac2=vka2b=wka3b3c3=uvwk3abc=(uvw)13kabc=kwk(u−1)+uk(v−1)+vk(w−1)+k=0uwk−wk+uvk−uk+vwk−vk+k=0
Commented by Prithwish sen last updated on 28/May/19
great thinking sir.
greatthinkingsir.
Commented by tanmay last updated on 28/May/19
k(uw+uv+vw−u−v−w)=0 k≠0 uw+uv+vw−u−v−k=0 u+v+k=uv+vw+uw wait...
k(uw+uv+vw−u−v−w)=0k≠0uw+uv+vw−u−v−k=0u+v+k=uv+vw+uwwait…
Answered by MJS last updated on 29/May/19
let b=pa ∧ c=qa (1) ⇒ (3−2q)p^2 +(3q^2 −3q−2)p+q(3−2q)=0 (2) ⇒ (1−q)p^2 +(q^2 +q−1)p+q(1−q)=0 ⇒ q=.198062 p= { ((.307979)),((.643104)) :} q=1.55496 p= { ((.307979)),((5.04892)) :} q=3.24698 p= { ((.643104)),((5.04892)) :} in all cases we get the result 25 btw (1/(.198...))=5.04...; (1/(.307...))=3.24...; (1/(.643...))=1.55...
letb=pa∧c=qa(1)⇒(3−2q)p2+(3q2−3q−2)p+q(3−2q)=0(2)⇒(1−q)p2+(q2+q−1)p+q(1−q)=0⇒q=.198062p={.307979.643104q=1.55496p={.3079795.04892q=3.24698p={.6431045.04892inallcaseswegettheresult25btw1.198…=5.04…;1.307…=3.24…;1.643…=1.55…

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *