Q1). If x/(b-c)(b+c-2a) = y/(c-a)(c+a-2b) = z/(a-b)(a+b-2c) then find the value of x+y+z = ?

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.

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