Test the proposed identity. Let A=B=C=60° so cosA=cosB=cosC=½, a=b=c (equilateral triangle).
2bccosA=a2=2accosB; a2+b2+c2=3a2, so the proposition is false, because a2≠3a2.
Cosine Rule:
a2=b2+c2-2bccosA; b2=a2+c2-2accosB; c2=a2+b2-2abcosC.
2bccosA=b2+c2-a2; 2accosB=a2+c2-b2,
2bccosA+2accosB=b2+c2-a2+a2+c2-b2=2c2,
bccosA+accosB=c2 is a true identity.
Also:
a2+b2+c2=b2+c2-2bccosA+a2+c2-2accosB+a2+b2-2abcosC,
a2+b2+c2=2(a2+b2+c2)-2(bccosA+accosB+abcosC),
2(bccosA+accosB+abcosC)=a2+b2+c2 is a true identity and is probably the one intended in the question.