Question

Given If x/b+c-a=y/c+a-b=z/a+b-c then proove that x(by+cz-ax)=y(cz+ax-by)=z(ax+by-cz) ​

Answer 1

Answer:

Explanation:

What would be a^x=b^y=c^z and ABC=1? How would I prove that xy+yz+zx=0?

Hi Guys, I am new to Quora..I love the enviroment here …. This answer can be done by taking logs as well ..However guys I have done it in the most simple way possible ..Feel free to ask me anything if you don't understand .

Let a^x=b^y=c^z=k

Therefore,

a=k^(1/x)

b=k^(1/y)

c=k^(1/z)

As abc=1

k^(1/x).k^(1/y).k^(1/z)=1

k^(1/x+1/y+1/z)= 1 or k^0( as any number to the power 0 is 1)

Comparing the powers we get

1/x+1/y+1/z=0

Or

(yz+xz+xy)/xyz=0

Or xy+yz+zx=0 hence proved :)

Answer 2

Answer:

Since all the terms are equal it must be equal to some constant value. So say

x/(b+c) = y/(c+a) = z/(a+b) =p

equating separately we get

x=p(b+c)

y=p(c+a)

z=p(a+b)

Now consider the equation

(b-c)x+(c-a)y+(a-b)z=k

k=p(b-c)(b+c)+p(c-a)(c+a)+p(a-b)(a+b)

k=p(b²-c²+c²-a²+a²-b²)

k=p(0)

k=0

So as considered

(b-c)x +(c-a)y+(a-b)z=0

Explanation: