a,b, c satisfy the following equations if $$ ac-b^2 $$ ≠ 0.
$$ ax+by=1 $$ , $$ bx+cy=1 $$ , $$ cx+ay=1 $$
Prove: $$ a^2+b^2+c^2=ab+bc+ac $$

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Uniteasy Admin · Accepted Answer

Solve the equations in group
$$ \begin{cases} ax+by=1& \ bx+cy=1 &\end{cases} $$
$$ \because ac-b^2 $$ ≠ 0, we get
$$ x=\dfrac{c-b}{ac-b^2} $$
$$ y=\dfrac{a-b}{ac-b^2} $$
Plug x, y in the third equation
$$ c \dfrac{c-b}{ac-b^2}+a\dfrac{a-b}{ac-b^2}=1 $$
Remove the denominator, we get
$$ a^2+b^2+c^2=ab+bc+ac $$

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