Let x/[(b-c)(b+c-2a)] = y/[(c-a)(c+a-2b)] = z/[(a-b)(a+b-2c)] = k
Now,
x/[(b-c)(b+c-2a)] = k
⇛x/[(b-c)(b+c-2a)] = k/1
By doing cross multiplication, we get
⇛1(x) = k(b-c)(b+c-2a)
⇛x = k(b²-c²-2a(b-c)
⇛x = k(b²-c²+2ac-2ab) --------Eqⁿ(1)
Also,
y/[(c-a)(c+a-2b)] = k
⇛y/[(c-a)(c+a-2b)] = k/1
⇛1(y) = k(c-a)(c+a-2b)]
⇛y = k(c²-a²-2bc+2ba) --------Eqⁿ(2)
Again,
z/[(a-b)(a+b-2c)] = k
⇛z/[(a-b)(a+b-2c)] = k/1
⇛1(z) = k(a-b)(a+b-2c)
⇛z = k(a-b)(a+b-2c)
⇛z = k(a²-b²-2ca+2bc) --------Eqⁿ(3)
Now,
Adding equation (1), (2) and (3), we get
x+y+z
⇛k(b²-c²+2ac-2ab + c²-a²-2bc+2ba + a²-b²-2ca+2bc)
⇛k × 0
⇛0.
Answer:-The value of x+y+z is 0.
