Question

If \[ a + b+ c = 0 \] then find the value of \[ {a^2 \over bc} + {b^2 \over ca} + {c^2 \over ab} ]\

Answer 1

\[{a ^{2} \over bc} + {b ^{2} \over ac} + {c ^{2} \over ab} \]

Answer 2

\[=\frac {a^3+b^3+c^3}{abc}\]

Answer 3

So the answer is 3 right?

Answer 4

I got 3 too.

Answer 5

ab(a+b+c) = a^2b+ab^2+abc =0 and other 2 cyclic
(a+b+c)^3 = a^3 + b^3 +c^3 + 6abc + (terms equal to above including 3abc) so
3abc/abc = 3

Answer 6

i dont understand how ?

Answer 7

If you expand (a+b+c)^3 completely you can equate all terms to 0 except 3abc

Answer 8

ok i follow that,
but i can not see how to get (a+b+c)^3

Answer 9

\[
0=\frac{(a+b+c)^3}{a b c}=\\
\frac{a^2}{b c}+\frac{b^2}{a
c}+\frac{c^2}{a b}+\frac{3
a}{b}+\frac{3 b}{a}+\frac{3
a}{c}+\frac{3 c}{a}+\frac{3
c}{b}+\frac{3 b}{c}+6\\

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

\frac{3 (-c) (-b) (-a)}{a b
c} = -\frac{a^2}{b
c}-\frac{b^2}{a
c}-\frac{c^2}{a b} \\

-3 = -\frac{a^2}{b
c}-\frac{b^2}{a
c}-\frac{c^2}{a b} \\

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

\]

Answer 10

(a+b+c)^3 = a^3+b^3+c^3 - 3abc (after zeroing other terms per my post above) so
a^3+b^3+c^3 = (a+b+c)^3 +3abc

The posters original question if you add it together gives
(a^3+b^3+c^3)/abc

OK?

Answer 11

now i see , thanks eliassaab, estudier ,
\[\frac{(a^3+b^3+c^3)}{abc}=\frac{(a+b+c)^3 +3abc}{abc}=3\]

Answer 12

OK, the numerator in the rhs comes from equating to zero all the other terms in the expansion of (a+b+c)^3...