Question

If a+b+c= 0,
then find the value of
(a+b-c)^3 + (b+c-a)^3 + (c+a-b)^3

Answer 1

I don't know if this helps much but:
If a+b+c=0,
then a+b=-c
and a+c=-b
and b+c=-a

So, (a+b-c)^3 = (-c-c)^3=-8c^3 by substitution "a+b+ with the above
(b+c-a)^3=(-a-a)^3=-8a^3
(c+a-b)^3=(-b-b)^3=-8b^3

=> the expression becomes -8(a^3+b^3+c^3)

Answer 2

I'm not sure what else to do

Answer 3

What class is this for?

Answer 4

Is there more information

Answer 5

If a, b, and c equal 0, then no matter what you do to it, it will always be 0 right?

Answer 6

^It says a+b+c=0, not a=b=c=0, so you could have like a=-1,b=1,c=0 or a=-2,b=-1,c=3, etc.

Answer 7

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+ b^2 + c^2 - ab -bc -ca)

You may use this :

Answer 8

did u get the answer?

Answer 9

no..

Answer 10

why not ..where are u stuck?

Answer 11

i m not getting ..i m stuck how 2 start

Answer 12

use the property @shubhamsrg told ..its really easy after that

Answer 13

kk

Answer 14

i m not getting it..

Answer 15

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+ b^2 + c^2 - ab -bc -ca).
How to use this.

Answer 16

a+b+c =0
make that substitution on RHS

Answer 17

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+ b^2 + c^2 - ab -bc -ca).
a^3 + b^3 + c^3 - 3abc = (0)(a^2+ b^2 + c^2 - ab -bc -ca)
a^3 + b^3 + c^3 - 3abc = 0

Answer 18

a^3 + b^3 + c^3 = 3abc

Answer 19

vat after that

Answer 20

substitute that into @kirbykirby 's work.

Answer 21

=> the expression becomes= -8(a^3+b^3+c^3)
=> the expression becomes= -8(3abc)=-24abc

Answer 22

is that answer.

Answer 23

Yep, seems right.

Answer 24

thanks