Practicing Success
If x = a (b - c) y = b (c - a) z = c (a - b) then find the value of \(\left[\frac{(x+y)\;(y+z)\;(z+x)}{4xyz}\right]\) |
\(\frac{1}{4}\) -\(\frac{1}{4}\) -\(\frac{1}{2}\) -\(\frac{1}{3}\) |
-\(\frac{1}{4}\) |
x = ab - ac y = bc - ba z = ca - cb ⇒ x + y + z = ab - ac + bc - ba + ca - cb = 0 So, ⇒ x + y = -z ⇒ y + z = -x ⇒ z + x = -y Put the values: ⇒ \(\left[\frac{(x+y)\;(y+z)\;(z+x)}{4xyz}\right]\) = \(\frac{(-x)×(-y)×(-z)}{4xyz}\) = -\(\frac{1}{4}\) |