Question

If $\frac{x}{(b-c)(b+c-2a)}=\frac{y}{(c-a)(a+b-2b)}=\frac{z}{(a-b)(a+b-2c)}$, then the value of (x+ y+ z)is _______.

• A. a+b+c
• B. 0
• C. $a^2+b^2+c^2$
• D. can't be determined

Answer 1

Answer: (B)

0

As the given rations or fractions are equal,
Let $x=\left[\right(b-c\left)\right(b+c-2a\left)\right]k$
$y=\left[\right(c-a\left)\right(c+a-2b\left)\right]k$
$z=\left[\right(a-b\left)\right(a+b-2c\left)\right]k$
$x+y+z=k\times \left[\right(b-c\left)\right(b+c-2a)+(c-a\left)\right(c+a-2b)+(a-b\left)\right(a+b-2c\left)\right]$
On expanding each product and adding, we get
$x+y+z=k\times 0=0$