Let $$ ax+by=1 $$ , $$ bx+cy=1 $$ , $$ cx+ay=1 $$ , and $$ ac-b^2≠ $$ 0. Prove the following expression is true:
$$ a^2+b^2+c^2=ab+bc+ac $$

Question

Let  $$ ax+by=1 $$ , $$ bx+cy=1 $$ , $$ cx+ay=1 $$ , and $$ ac-b^2≠ $$ 0. Prove the following expression is true:
$$ a^2+b^2+c^2=ab+bc+ac $$

Uniteasy Admin · Accepted Answer

$$ \because ax+by=1 $$
$$ 	herefore abx+b^2y = b $$n
$$ \because bx+cy=1 $$
$$ 	herefore abx+acy = a $$n
Subtract (2) from (1), 
$$ (b^2-ac)y = b-a $$
 
$$ 	herefore  y = \dfrac{b-a}{b^2-ac} $$n
Similarly,
$$ \because ax+by=1 $$
$$ 	herefore acx+bcy=c $$n
$$ \because bx+cy=1 $$ 
$$ 	herefore  b^2x+bcy=b $$n
Subtract (5) from (4),
$$ (ac-b^2)x = c-b $$
$$ 	herefore x=\dfrac{b-c}{b^2-ac} $$n
Plug the expressions of (3) and (6) in 
$$ cx+ay=1 $$
We get,
$$ a^2+b^2+c^2=ab+bc+ac $$
Now we have proved the expression.

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