If \(\dfrac{1}{2b+2c}+\dfrac{1}{2c+2a}=\dfrac{1}{a+b}\), then what is \(a^2+b^2+c^2\) equal to?
\(3c^2\)
The equation simplifies to \(\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2}{a+b}\). Combining the left side: \(\dfrac{a+b+2c}{(b+c)(c+a)}=\dfrac{2}{a+b}\). Cross-multiplying: \((a+b+2c)(a+b)=2(b+c)(c+a)\). Expanding both sides and cancelling the common term \(2c(a+b)\) gives \((a+b)^2=2ab+2c^2\), i.e. \(a^2+b^2=2c^2\). Therefore \(a^2+b^2+c^2=2c^2+c^2=3c^2\).
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