<s>[INST] equilateral triangle; $3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$ In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$
Please prove this identity.
source: http://www.qbyte.org/puzzles/p117s.html
 [/INST] Here's a brute-force approach:

Abbreviating the measures of angles surrounding point $P$ thusly,
$$\alpha := |\angle BPC| \qquad \beta := |\angle CPA| \qquad \gamma := |\angle APB|$$
we have 
$$\cos\gamma = \cos\left(2\pi - \alpha - \beta\right) = \cos\left(\alpha+\beta\right) = \cos\alpha \cos \beta - \sin\alpha \sin\beta $$
so that
$$\left( \cos\alpha \cos\beta - \cos\gamma \right)^2 = \left(1-\cos^2\alpha\right)\left(1-\cos^2\beta\right)$$
Now, substitute-in these relations from the Law of Cosines
$$\cos\alpha = \frac{b^2+c^2-d^2}{2bc} \qquad
\cos\beta = \frac{c^2+a^2-d^2}{2ca} \qquad
\cos\gamma = \frac{a^2+b^2-d^2}{2ab}$$
and do a bit of algebra.
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