Question

let x be a real number if a= 2011x+9997 b=2011x+9998 c=2011x+9999 Find the value of a²+b²+c²‐ab-bc-ca​

Answer 1

Answer:

The answer is 3

Step-by-step explanation:

Given  

             $a=2011x+9997,b=2011x+9998,c=2011x+9999$

Clearly $a-b=-1, b-c=-1,c-a=2$

Therefore  

                   $a^2+b^2+c^2-ab-bc-ca$

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

             $=a(a-b)+b(b-c)+c(c-a)\\=a(-1)+b(-1)+c(2)$  using above information

             $=2c-(a+b)=(c-a)+(c-b)\\=2+1=3$

Answer 2

Step-by-step explanation:

Enclosure contains the solution.