If a2 + b2 + c2 + 3 = 2 (a – b – c), then the value of 2a – b + c is:
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Value of 2a-b+c=1
Step-by-step explanation:
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
a=b=c=1
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