x2+2x(a+b+c)+(a+b+c)2=(x+a+b+c)2, therefore, if (a+b+c)2=3(ab+bc+ca), then we have a perfect square.
(a+b+c)2=((a+b)2+2c(a+b)+c2=a2+2ab+b2+2ac+2bc+c2, that is:
a2+b2+c2+2(ab+bc+ca), which has to equal 3(ab+bc+ca).
Therefore a2+b2+c2=ab+bc+ca. There are three terms on each side so let's equate them:
a2=ab, b2=bc, c2=ac⇒a=b, b=c, c=a, that is, a=b=c (a,b,c≠0).
Are there any other solutions? Suppose a2=bc, b2=ca, c2=ab, then:
a2-b2=bc-ca=c(b-a), (a-b)(a+b)=c(b-a)⇒a=b or c=-(a+b), or a+b+c=0;
b2-c2=(b-c)(b+c)=ca-ab=a(c-b)⇒b=c or a=-(b+c), or a+b+c=0;
c2-a2=(c-a)(c+a)=ab-bc=b(a-c)⇒a=c or b=-(a+c), or a+b+c=0.
There may be another solution: a+b+c=0 or c=-(a+b). If we substitute for c in x2+2x(a+b+c)+(a+b+c)2, we get x2, which is a perfect square.
Therefore there are two solutions: a=b=c or a+b+c=0.