2(a+b)(b+c)(a+c)+abc
=2[a^2+a(b+c)+bc](b+c）+abc
=2(b+c)a^2+a(2b^2+5bc+2c^2)+2bc(b+c),
△（a)=(2b^2+5bc+2c^2)^2-16bc(b+c)^2
=[2(b+c)^2+bc]^2-16bc(b+c)^2
=[2(b+c)^2-bc]^2-8bc(b+c)^2,不是平方式,
所以原式在有理数范围内不能分解因式.
(a+b)(b+c)(a+c)+abc
=[a^2+a(b+c)+bc](b+c）+abc
=(b+c)a^2+a(b^2+3bc+c^2)+bc(b+c),
△‘=[(b+c)^2+bc]^2-4bc(b+c)^2
=[(b+c)^2-bc]^2,
a=-bc/(b+c),或a=-(b+c),
∴(a+b)(b+c)(a+c)+abc
=（b+c)[a+bc/(b+c)](a+b+c)
=(ab+bc+ca)(a+b+c).