Question

If a, b, c are all non -zero and a+b+c= 0, prove a²÷ bc +b²÷ ac +c²÷ ab =3

Answer 1

Combining the fractions gives

$\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}$

Since $a+b+c=0$, we also have

$(a+b+c)^3=0$

$a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc=0$

If we add and subtract the last 7 terms of the left hand side to/from the numerator, we get

$\dfrac{(a+b+c)^3-(3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc)}{abc}$

$=-3\dfrac{a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+2abc}{abc}$

$=-3\dfrac{(a+b)(a+c)(b+c)}{abc}$

Also because $a+b+c=0$, we have

$\begin{cases}a=-(b+c)\\b=-(a+c)\\c=-(a+b)\end{cases}$

so we find that

$-3\dfrac{(a+b)(a+c)(b+c)}{abc}=-3\dfrac{(-c)(-b)(-a)}{abc}=3\dfrac{abc}{abc}=3$