Question

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

pls solve and give the explanation​

Answer 1

Answer:

$\frac{a^2 + b^2 +c ^2}{a*b + a*c + b*c} = -2$

Step-by-step explanation:

if:

a + b + c = 0

then we can rewrite:

c = -a - b

Now we can replace this in our expression:

$\frac{a^2 + b^2 + c^2}{a*b + a*c + b*c} = \frac{a^2 + b^2 + (-a - b)^2}{a*b + a*(-a - b) + b*(-a - c)}$

Now we can rewrite:

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

Then we can rewrite our expression as:

$\frac{a^2 + b^2 + a^2 + b^2 + 2*a*b}{a*b - a^2 - a*b - b^2 - b*a}$

Let's keep simplifying this:

$\frac{2*(a^2 + b^2 + a*b)}{-a*b - b^2 - a^2}$

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).

$\frac{2*(a^2 + b^2 + a*b)}{-a*b - b^2 - a^2} = \frac{2*(a^2 + b^2 + a*b)}{-(a^2 + b^2 + a*b)} = -2$

Then the expression is equal to -2

$\frac{a^2 + b^2 +c ^2}{a*b + a*c + b*c} = -2$