Question

If a , b, c are all non zero and a+ b + c=0 prove that: a^2/bc + b^2/ca + c^2/ab = 3

Answer 1

Please find attached photograph for your answer

Answer 2

`(a^2)/(bc) + (b^2)/(ca) + (c^2)/(ab) = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |* abc \\ \\ (a^2 * abc)/(bc) + (b^2 * abc)/(ca) + (c^2 * abc)/(ab)=3abc \\ \\ a^2 * a + b^2 * b + c^2 * c = 3abc \\ \\ a^3+b^3+c^3=3abc \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |-3abc \\ \\ a^3+b^3+c^3-3abc=0 \\ \\`

You know the value of a+b+c, so try to factor a+b+c out of the expression:


`a^3+b^3+c^3-3abc= \\ \\ a^3+a^2b+a^2c+ab^2+b^3+b^2c+ac^2+bc^2+c^3-a^2b-ab^2-abc \\ -abc-b^2c-bc^2-a^2c-abc-ac^2= \\ \\ a^2(a+b+c)+b^2(a+b+c)+c^2(a+b+c)-ab(a+b+c) \\ -bc(a+b+c)-ac(a+b+c)= \\ \\ (a^2+b^2+c^2-ab-bc-ac)(a+b+c)`

Back to the equation:


`(a^2+b^2+c^2-ab-bc-ac)(a+b+c)=0`

You know that a+b+c=0.


`(a^2+b^2+c^2-ab-bc-ac) * 0=0 \\ \\ 0=0 \\ \\ LHS=RHS`

The equation
`(a^2)/(bc) + (b^2)/(ca) + (c^2)/(ab) =3` is true for any values of a, b, c, such that a+b+c=0. Proved.