Question

If a+b+c=0 then find the value of (a/b-c +b/c-a +c/a-b)(a-b/c+b-c/a +c-a/b)​

Answer 1

Answer:

As a+b+c = 0 , a³+b³+c³ = 3abc...…..(2)

Let x be (a-b)/c , z be (c-a)/b and y be (b-c)/a  

Substituting it in the above equation:

= (1/x+1/y+1/z) (x+y+z)

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

=3+(y+z)/x+(x+z)/y+(x+y)/z...…..(1)

Now lets find the value of (y+z)/x = 1/x (y+z)

= c/(a-b) ((b-c)/a+(c-a)/b)  

= c/(a-b) ((b²-bc+ac-a²)/(ab))  

= c/(a-b) (((ac-bc)-(a²-b²))/(ab))  

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

= c/(a-b) ((a-b)(c-(a+b))/(ab))  

= c(c-(a+b))/(ab)  

= c((c-a-b+c-c)/(ab))    (adding and subtracting by c)

= c (2c-(a+b+c)/(ab))    (a+b+c=0)

= c(2c/ab)

= 2c²/ab

Similarly the values of (z+x)/y=2a²/bc and (x+y)/z=2b²/ca

Substituting in eq 1

= 3+(2c²/ab+2a²/bc+2b²/ca)  

= 3+2(c³+a³+b³)/abc

= 3+2(3abc)/abc …. from (2)  

= 3+6

= 9  

Hence proved.

Step-by-step explanation: