Question

If a+b+c=0 then prove that (a+b-c)^3+(c+a-b)^3+(b+c-a)^3= -24abc

Answer 1

Answer:

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Step-by-step explanation:

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. 

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Answer 2

Step-by-step explanation:

We know that

if a + b + c = 0

then a³ + b³ + c³ = 3abc

Now ,

(a+b-c)³ + (c+a-b)³ + (b+c-a)³

= (-c-c)³ + (-b-b)³ + (-a-a)³

= (-2c)³ + (-2b)³ + (-2a)³

= - 8c³ - 8b³ - 8a³

= -8 (a³ + b³ + c³)

= -8(3abc)

= -24abc

Hence proof