Question

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

Answer 1

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. 
 

Answer 2

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

$LHS\\= (\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} ) * (\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c} )\\\\multiply\ the\ terms\\\\=\frac{a}{b-c}\frac{b-c}{a}+\frac{a}{b-c}\frac{c-a}{b}+\frac{a}{b-c}\frac{a-b}{c}+\frac{b}{c-a}\frac{b-c}{a}+\frac{b}{c-a}\frac{c-a}{b}\\\\+\frac{b}{c-a}\frac{a-b}{c}+\frac{c}{a-b}\frac{b-c}{a}+\frac{c}{a-b}\frac{c-a}{b}+\frac{c}{a-b}\frac{a-b}{c}$

$=1+\frac{a}{b-c}\frac{c-a}{b}+\frac{a}{b-c}\frac{a-b}{c}+\frac{b}{c-a}\frac{b-c}{a}+1+\frac{b}{c-a}\frac{a-b}{c}\\\\ +\frac{c}{a-b}\frac{b-c}{a}+\frac{c}{a-b} \frac{c-a}{b} +1\\\\=3+\frac{(c-a)ac+ab(a-b)}{(b-c)bc}+\frac{bc(b-c)+ba(a-b)}{(c-a)ac}+\frac{bc(b-c)+ac(c-a)}{(a-b)ab}$

$=3+\frac{ac^2-a^2c+a^2b-ab^2}{(b-c)bc}+\frac{b^2c-bc^2+ba^2-b^2a}{(c-a)ac}+\frac{b^2c-bc^2+ac^2-a^2c}{(a-b)ab}\\\\=3+\frac{(a^2-ab-ac)(b-c)}{(b-c)bc}+\frac{(c-a)(b^2-bc-ab)}{(c-a)ac}+\frac{(c^2-bc-ac)(a-b)}{(a-b)ab}$


$=3+\frac{(a^2-ab-ac)}{bc}+\frac{(b^2-bc-ab)}{ac}+\frac{(c^2-bc-ac)}{ab}\\\\=3+\frac{(a^3-a^2b-a^2c)}{abc}+\frac{(b^3-b^2c-ab^2)}{abc}+\frac{(c^3-bc^2-ac^2)}{abc}\\\\=3+\frac{a^3+b^3+c^3}{abc}+\frac{-a^2(b+c)-b^2(c+a)-c^2(a+b)}{abc}\\\\=3+\frac{3abc}{abc}+\frac{-a^2(-a)-b^2(-b)-c^2(-c)}{abc}\\\\=3+3+\frac{a^3+b^3+c^3}{abc}\\\\=9$

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