Question

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

Answer 1

Please find attached photograph for your answer

Answer 2

Combining the fractions gives

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

`((a+b+c)^3-(3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc))/(abc)`

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

`=-3((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((a+b)(a+c)(b+c))/(abc)=-3((-c)(-b)(-a))/(abc)=3(abc)/(abc)=3`