Question

a2 + b2 + 2c in computer​

Answer 1

Given a2+b2+c2+3=2(a+b+c)

Expanding RHS we get a2+b2+c2+3=2a+2b+2c

Carrying RHS to LHS -> a2+b2+c2+3-2a-2b-2c=0

We have a2-2a, so we try if any identity can be applied here - a2-2a+1 is the expansion for (a-1)2. Since we have three variables a,b,c the constant here 3 is spilt as 1 1 1

a2-2a+1 + b2-2b+1 c2-2c+1 = 0

Applying the identity, (a−b)2=a2- 2ab + b2 we get

(a−1)2+(b−1)2+(c−1)2=0

ie; all the three terms shd be 0

so (a−1)2=(b−1)2=(c−1)2=0

Answer 2

Answer:

a×2 + b×2 + 2×c

Explanation:

Hope you that the answer is clear to you