A problem of algebra!

Algebra Level 2

Let a a , b b and c c be such that a + b + c = 0 a+b+c=0 and

P = a 2 2 a 2 + b c + b 2 2 b 2 + c a + c 2 2 c 2 + a b P=\dfrac{a^{2}}{2a^{2}+bc}+\dfrac{b^{2}}{2b^{2}+ca}+\dfrac{c^{2}}{2c^{2}+ab} is defined.

What is the value of P P ?

2 4 1 3

S P by s p , 16, India

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3 solutions

S P
May 12, 2018

a 2 2 a 2 + b c = a 2 a 2 + a ( − b − c ) + b c = a 2 ( a − b ) ( a − c ) \begin{aligned}\dfrac{a^2}{2a^2+bc}=\dfrac{a^2}{a^2+a(-b-c)+bc} \\& =\dfrac{a^2}{(a-b)(a-c)}\end{aligned} [as a + b + c = 0 a+b+c=0 then a = − b − c a=-b-c ]

Similarly, b 2 ( b − c ) ( b − a ) \dfrac{b^2}{(b-c)(b-a)} and c 2 ( c − a ) ( c − b ) \dfrac{c^2}{(c-a)(c-b)}

Therefore P = a 2 ( a − b ) ( a − c ) + b 2 ( b − c ) ( b − a ) + c 2 ( c − a ) ( c − b ) = − a 2 ( b − c ) − b 2 ( c − a ) − c 2 ( a − b ) ( a − b ) ( b − c ) ( c − a ) \begin{aligned}P=\dfrac{a^2}{(a-b)(a-c)}+\dfrac{b^2}{(b-c)(b-a)}+\dfrac{c^2}{(c-a)(c-b)} \\&= \dfrac{-a^2(b-c)-b^2(c-a)-c^2(a-b)}{(a-b)(b-c)(c-a)} \end{aligned}

Now add + a b c − a b c +abc-abc to the numerator and by solving we get

( a − b ) ( b − c ) ( c − a ) ( a − b ) ( b − c ) ( c − a ) = 1 \dfrac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=\boxed{1}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Hana Wehbi
May 11, 2018

I posted the same problem. It is called ISI (1)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Qyu Zhang
May 11, 2018

A stupid way: Let a=1, b=2, c=-3, and then, you know...

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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