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

Collected in the board: Algebraic equation

Steven Zheng posted 3 months ago


Answer

\because ax+by=1

\therefore abx+b^2y = b
(1)
\because bx+cy=1

\therefore abx+acy = a
(2)

Subtract (2) from (1),

(b^2-ac)y = b-a

\therefore y = \dfrac{b-a}{b^2-ac}
(3)

Similarly,

\because ax+by=1

\therefore acx+bcy=c
(4)
\because bx+cy=1

\therefore b^2x+bcy=b
(5)

Subtract (5) from (4),

(ac-b^2)x = c-b

\therefore x=\dfrac{b-c}{b^2-ac}
(6)

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.

Steven Zheng posted 3 months ago

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