Question

If a +b+c = 0 then the value of a²+b²+c²/bc+ca+ab

pls solve and give the explanation​

Answer 1

`(a^2 + b^2 +c ^2)/(a*b + a*c + b*c) = -2`

Explanation:

if:

a + b + c = 0

then we can rewrite:

c = -a - b

Now we can replace this in our expression:

`(a^2 + b^2 + c^2)/(a*b + a*c + b*c) = (a^2 + b^2 + (-a - b)^2)/(a*b + a*(-a - b) + b*(-a - c))`

Now we can rewrite:

(-a - b)^2 = a^2 + 2*a*b + b^2

Then we can rewrite our expression as:

`(a^2 + b^2 + a^2 + b^2 + 2*a*b)/(a*b - a^2 - a*b - b^2 - b*a)`

Let's keep simplifying this:

`(2*(a^2 + b^2 + a*b))/(-a*b - b^2 - a^2)`

Here is easy to see the simplification we need to do, as we have the same factor in the denominator as in the numerator (just a different of sign).

`(2*(a^2 + b^2 + a*b))/(-a*b - b^2 - a^2) = (2*(a^2 + b^2 + a*b))/(-(a^2 + b^2 + a*b)) = -2`

Then the expression is equal to -2

`(a^2 + b^2 +c ^2)/(a*b + a*c + b*c) = -2`