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-

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 = \frac{a^{2}}{2 a^{2} + b c} + \frac{b^{2}}{2 b^{2} + c a} + \frac{c^{2}}{2 c^{2} + a b}

is defined . What is the value of P ?

Commented by dioph last updated on 27/Aug/17

P = \frac{12 a^{2} b^{2} c^{2} + 4 a^{3} b^{3} + 4 a^{3} c^{3} + 4 b^{3} c^{3} + a^{4} b c + {a b}^{4} c + {a b c}^{4}}{9 a^{2} b^{2} c^{2} + 4 a^{3} b^{3} + 4 a^{3} c^{3} + 4 b^{3} c^{3} + 2 a^{4} b c + 2 {a b}^{4} c + 2 {a b c}^{4}}

{(a + b + c)}^{3} = a^{3} + b^{3} + c^{3} + 3 a^{2} b + 3 {a b}^{2} + 3 {a c}^{2} + 3 {b c}^{2} + 3 a^{2} c + 3 b^{2} c + 6 a b c

a^{3} + b^{3} + c^{3} + 6 a b c = - 3 a b (a + b) - 3 a c (a + c) - 3 b c (b + c)

a^{3} + b^{3} + c^{3} + 6 a b c = 3 a b c + 3 a b c + 3 a b c

a^{3} + b^{3} + c^{3} + 6 a b c = 9 a b c

a^{3} + b^{3} + c^{3} = 3 a b c (*)

P = \frac{4 (a^{3} b^{3} + a^{3} c^{3} + b^{3} c^{3}) + a b c (a^{3} + b^{3} + c^{3} + 12 a b c)}{4 (a^{3} b^{3} + a^{3} c^{3} + b^{3} c^{3}) + a b c (2 a^{3} + 2 b^{3} + 2 c^{3} + 9 a b c)}

P = \frac{4 (a^{3} b^{3} + a^{3} c^{3} + b^{3} c^{3}) + a b c (15 a b c)}{4 (a^{3} b^{3} + a^{3} c^{3} + b^{3} c^{3}) + a b c (15 a b c)}

P = 1

Answered by Tinkutara last updated on 27/Aug/17

Since a + b + c = 0,

∴ a = - b - c \Rightarrow a^{2} = - a b - a c

2 a^{2} + b c = a^{2} - a b - a c + b c = (a - b) (a - c)

Similarly, 2 b^{2} + c a = (b - a) (b - c),

2 c^{2} + a b = (c - a) (c - b)

Hence,

P = \frac{- a^{2}}{(a - b) (c - a)} + \frac{- b^{2}}{(a - b) (b - c)} + \frac{- c^{2}}{(c - a) (b - c)}

= \frac{- \sum_{c y c} 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 .

Answered by ajfour last updated on 27/Aug/17

I f a + b + c = 0,

L e t a = p c, b = q c

a + b = - c \Rightarrow p + q = - 1

P = \frac{p^{2} c^{2}}{2 p^{2} c^{2} + {q c}^{2}} + \frac{q^{2} c^{2}}{2 q^{2} c^{2} + {p c}^{2}} + \frac{c^{2}}{2 c^{2} + {p q c}^{2}}

= \frac{p^{2}}{2 p^{2} + q} + \frac{q^{2}}{2 q^{2} + p} + \frac{1}{2 + p q}

= \frac{4 p^{2} q^{2} + p^{3} + q^{3}}{4 p^{2} q^{2} + 2 (p^{3} + q^{3}) + p q} + \frac{1}{2 + p q}

= \frac{4 p^{2} q^{2} - (p^{2} + q^{2} - p q)}{4 p^{2} q^{2} - 2 (p^{2} + q^{2} - p q) + p q} + \frac{1}{2 + p q}

= \frac{4 p^{2} q^{2} - (1 - 3 p q)}{4 p^{2} q^{2} - 2 (1 - 3 p q) + p q} + \frac{1}{2 + p q}

l e t p q = r

P = \frac{4 r^{2} + 3 r - 1}{4 r^{2} + 7 r - 2} + \frac{1}{2 + r}

= 1 + \frac{(1 - 4 r)}{4 r^{2} + 7 r - 2} + \frac{1}{2 + r}

= 1 + \frac{(2 - 7 r - 4 r^{2}) + (4 r^{2} + 7 r - 2)}{(4 r^{2} + 7 r - 2) (2 + r)}

\Rightarrow P = 1 + 0 .

Commented by ajfour last updated on 27/Aug/17

g o o d q u e s t i o n, t h a n k s .

Commented by Tinkutara last updated on 27/Aug/17

Thank you very much Sir!