Question

can anyone solve this ....

Answer 1

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

On Multiplying  and dividing by 2, we get

= (a + b + c)/2 (2a^2 + 2b^2 + 2c^2 - 2ab - 2ca - 2bc). 

= (a + b + c)/2 (a^2 + b^2 - 2ab + b^2 +c^2- 2bc + c^2+b^2-2bc). 

= (a+b+c)/2( (a-b)^2 + (b-c)^2 + (c-a)^2 )

= 1/2(a+b+c)( (a-b)^2 + (b-c)^2 + (c-a)^2 )).


Hope this helps!