# If a+b+c=0,then prove that (a/b-c + b/c-a +c/a-b) (b-c/a + c-a/b + a-b/c) =9

2
by rahulnandi143
the question is nt applicble 4 al though. if u put a=-3, b=1, c=2. dn ans nt cumin 9

2015-05-04T09:53:48+05:30
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)
= 2c
²/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.

thanx
welcome :)
2015-05-04T13:02:16+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
We know that      a³ + b³ + c³ - 3abc= (a + b + c) (a² + b² + c² - ab - bc - ca)
If (a+b+c) = 0 ,   then       a³ + b³ + c³ = 3abc
Also  c  = - a - b

Refresh the browser, if the equations do not appear properly.

click on thanks azur blue button above please