Question

Factorise the following. :)

(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 + x=3a+b+c3x=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)

Hope it was Helpful

Mark it as Brainliest. Please

Answer 2

Step-by-step explanation:

x= a+b+c/3 (Given)

3x= a+b+c

Then, a= x-a

b=x-b

c=x-c

So, a+b+c= x-a+x-b+x-c

= x+x+x-a-b-c

= 3x-a-b-c

We can also write it like:-

3x= a+b+c

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

=>a3+b3+c3=3abc (Usingidentity)

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

Hope it helps..✌️