Question

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

Steven Zheng Steven Zheng posted 2 months ago


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

Steven Zheng Steven Zheng posted 2 months ago

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