Question

Factorise :

• (x-a)³+(x-b)³+(x-c)³ , where x = (a+b+c)/3

Answer 1

$x = \frac{a + b + c}{3} \\ \\ 3x = a + b + c$

Then a = x-a , b = x-b and c = x-c

So,

a+b+c =

x-a+x-b+x-c

3x -a-b-c

We can write 3x = a+b+c

=> a+b+c -a-b-c = 0

Identity to be used :

a³+b³+c³ = 3abc

= 3(x-a)(x-b)(x-c)

Answer 2

X = a + b + c / 3

3x = a + b + c

Then,

a = x-a ,

b = x-b ,

c = x-c ,

So, 

a+b+c = x-a+x-b+x-c 

3x -a-b-c 

We can also write it like :

3x = a+b+c 

> a+b+c -a-b-c

= 0 

Identity to be used : 

a³+b³+c³ = 3abc 

= 3(x-a)(x-b)(x-c) Answer