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Question Number 20430 by Tinkutara last updated on 26/Aug/17

Let a, b and c be such that a + b + c = 0  and  P = (a^2 /(2a^2  + bc)) + (b^2 /(2b^2  + ca)) + (c^2 /(2c^2  + ab))  is defined. What is the value of P?

Leta,bandcbesuchthata+b+c=0andP=a22a2+bc+b22b2+ca+c22c2+abisdefined.WhatisthevalueofP?

Commented by dioph last updated on 27/Aug/17

P = ((12a^2 b^2 c^2 +4a^3 b^3 +4a^3 c^3 +4b^3 c^3 +a^4 bc+ab^4 c+abc^4 )/(9a^2 b^2 c^2 +4a^3 b^3 +4a^3 c^3 +4b^3 c^3 +2a^4 bc+2ab^4 c+2abc^4 ))  (a+b+c)^3 =a^3 +b^3 +c^3 +3a^2 b+3ab^2 +3ac^2 +3bc^2 +3a^2 c+3b^2 c+6abc  a^3 +b^3 +c^3 +6abc=−3ab(a+b)−3ac(a+c)−3bc(b+c)  a^3 +b^3 +c^3 +6abc=3abc+3abc+3abc  a^3 +b^3 +c^3 +6abc=9abc  a^3 +b^3 +c^3 =3abc (∗)  P = ((4(a^3 b^3 +a^3 c^3 +b^3 c^3 )+abc(a^3 +b^3 +c^3 +12abc))/(4(a^3 b^3 +a^3 c^3 +b^3 c^3 )+abc(2a^3 +2b^3 +2c^3 +9abc)))  P = ((4(a^3 b^3 +a^3 c^3 +b^3 c^3 )+abc(15abc))/(4(a^3 b^3 +a^3 c^3 +b^3 c^3 )+abc(15abc)))  P = 1

P=12a2b2c2+4a3b3+4a3c3+4b3c3+a4bc+ab4c+abc49a2b2c2+4a3b3+4a3c3+4b3c3+2a4bc+2ab4c+2abc4(a+b+c)3=a3+b3+c3+3a2b+3ab2+3ac2+3bc2+3a2c+3b2c+6abca3+b3+c3+6abc=3ab(a+b)3ac(a+c)3bc(b+c)a3+b3+c3+6abc=3abc+3abc+3abca3+b3+c3+6abc=9abca3+b3+c3=3abc()P=4(a3b3+a3c3+b3c3)+abc(a3+b3+c3+12abc)4(a3b3+a3c3+b3c3)+abc(2a3+2b3+2c3+9abc)P=4(a3b3+a3c3+b3c3)+abc(15abc)4(a3b3+a3c3+b3c3)+abc(15abc)P=1

Answered by Tinkutara last updated on 27/Aug/17

Since a + b + c = 0,  ∴ a = − b − c ⇒ a^2  = − ab − ac  2a^2  + bc = a^2  − ab − ac + bc = (a − b)(a − c)  Similarly, 2b^2  + ca = (b − a)(b − c),  2c^2  + ab = (c − a)(c − b)  Hence,  P = ((−a^2 )/((a − b)(c − a))) + ((−b^2 )/((a − b)(b − c))) + ((−c^2 )/((c − a)(b − c)))  = ((−Σ_(cyc) a^2 (b − c))/((a − b)(b − c)(c − a)))  After expanding the denominator, we will find this  value just the same as numerator.  So, P = 1.

Sincea+b+c=0,a=bca2=abac2a2+bc=a2abac+bc=(ab)(ac)Similarly,2b2+ca=(ba)(bc),2c2+ab=(ca)(cb)Hence,P=a2(ab)(ca)+b2(ab)(bc)+c2(ca)(bc)=cyca2(bc)(ab)(bc)(ca)Afterexpandingthedenominator,wewillfindthisvaluejustthesameasnumerator.So,P=1.

Answered by ajfour last updated on 27/Aug/17

If a+b+c=0,  Let a=pc,  b=qc   a+b=−c ⇒   p+q=−1  P=((p^2 c^2 )/(2p^2 c^2 +qc^2 ))+((q^2 c^2 )/(2q^2 c^2 +pc^2 ))+(c^2 /(2c^2 +pqc^2 ))   =(p^2 /(2p^2 +q))+(q^2 /(2q^2 +p))+(1/(2+pq))   =((4p^2 q^2 +p^3 +q^3 )/(4p^2 q^2 +2(p^3 +q^3 )+pq))+(1/(2+pq))   =((4p^2 q^2 −(p^2 +q^2 −pq))/(4p^2 q^2 −2(p^2 +q^2 −pq)+pq))+(1/(2+pq))   =((4p^2 q^2 −(1−3pq))/(4p^2 q^2 −2(1−3pq)+pq))+(1/(2+pq))   let    pq=r   P = ((4r^2 +3r−1)/(4r^2 +7r−2))+(1/(2+r))       =1+(((1−4r))/(4r^2 +7r−2))+(1/(2+r))       =1+(((2−7r−4r^2 )+(4r^2 +7r−2))/((4r^2 +7r−2)(2+r)))  ⇒ P =1+0 .

Ifa+b+c=0,Leta=pc,b=qca+b=cp+q=1P=p2c22p2c2+qc2+q2c22q2c2+pc2+c22c2+pqc2=p22p2+q+q22q2+p+12+pq=4p2q2+p3+q34p2q2+2(p3+q3)+pq+12+pq=4p2q2(p2+q2pq)4p2q22(p2+q2pq)+pq+12+pq=4p2q2(13pq)4p2q22(13pq)+pq+12+pqletpq=rP=4r2+3r14r2+7r2+12+r=1+(14r)4r2+7r2+12+r=1+(27r4r2)+(4r2+7r2)(4r2+7r2)(2+r)P=1+0.

Commented by ajfour last updated on 27/Aug/17

good question, thanks.

goodquestion,thanks.

Commented by Tinkutara last updated on 27/Aug/17

Thank you very much Sir!

ThankyouverymuchSir!

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