Question

Prove that:
if
\[a + 2b + 3c = 7x\]
then

\[a^2 + b^2 + c^2 = (x-a)^2 + (2x-b)^2 + (3x-c)^2\]

Answer 1

a+2b+3c=7x
x=(a+2b+3c)/7
Put this into (x−a)^2+(2x−b)^2+(3x−c)^2
(x−a)^2+(2x−b)^2+(3x−c)^2
=(((a+2b+3c)/7)-a)^2 +(2(a+2b+3c)/7 -b)^2 +(3(a+2b+3c)/7 -c)^2
Then by expansion and grouping terms, we can get the answer.

Answer 2

That seems way too long though.

Answer 3

Here is how I did it