Question

resolve into factors:(a^3c+b^3a+bc^3-b^3c-c^3a-a^3b)​

Answer 1

Answer:

Let

f(a)=ab3−a3b+a3c−ac3+bc3−b3c.

If a=b, then f(a)=0. Therefore

(a−b)∣f(a). Analogously

(a−c)∣f(a); so,

(a−b)(a−c)∣f(a). But we can take it directly:

ab3−a3b+a3c−ac3+bc3−b3c

=ab(b2−a2)+c3(b−a)−c(b3−a3)=

=ab(b−a)(b+a)+c3(b−a)−c(b−a)(b2+ba+a2)=

=(b−a)[ab(b+a)+c3−cb2−cab−ca2]

ab(b+a)+c3−cb2−cab−ca2

=ab(a+b)+c(c2−a2)−cb(a+b)=

=(a−c)b(a+b)−c(a−c)(a+c)

=(a−c)[b(a+b)−c(a+c)]

And then

b(a+b)−c(a+c)=ba+b2−ca−c2

=a(b−c)+(b2−c2)=

=a(b−c)+(b+c)(b−c)=(b−c)(a+b+c).

So,

ab3−a3b+a3c−ac3+bc3−b3c

=(b−a)(a−c)(b−c)(a+b+c)