Question

by eliminating a b c from homogeneous equation x=a/b-c y=b/c-a z=c/a-b where a b c are not equal to zero then xy+yz+zx=​

Answer 1

Answer:

Step-by-step explanation:

XY + YZ+ ZX = (a / b-c ) (b /c-a)+ (b/c-a) (c/a-b)+ (c/ a-b)(a/b-c)

= ab (a-b) + bc(b-c) +ca(c-a) / (a-b) (b-c)(c-a)

= a²b -ab²+ b²c - bc²+c²a- a²c / (a-b) (b-c)(c-a)

= a²b -ab²+ b²c - bc²+c²a- a²c/ abc - a²b-ac²+a²c-b²c+ab²+bc²-abc

=a²b -ab²+ b²c - bc²+c²a- a²c /-a²b +ab²- b²c +bc²-c²a+ a²c

= a²b -ab²+ b²c - bc²+c²a- a²c/ -(a²b -ab²+ b²c - bc²+c²a- a²c)

= -1