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Factor  ab3a3b+bc3b3c+ca3c3a.

 Jun 4, 2020
 #1
avatar+659 
+2

Group every two terms and factor (a,b,c) respectively:

a(b3a2b)+b(c3b2c)+c(a3c2a)

 

Now factor (b,c,a) respectively:

ab(b2a2)+bc(c2b2)+ca(a2c2)

 

Now obviously follow up to factor the subtraction of two perfect squares.

ab(ba)(b+a)+bc(cb)(c+b)+ca(ac)(a+c)

 

Not sure if this can be factored further

 Jun 5, 2020
 #2
avatar+26396 
+4

Factor ab3a3b+bc3b3c+ca3c3a.

 

ab3a3b+bc3b3c+ca3c3a=a3b+ab3+ca3b3cc3a+bc3=ab(a2b2)+c(a3b3)c3(ab)=ab(ab)(a+b)+c(ab)(a2+ab+b2)c3(ab)=(ab)(ab(a+b)+c(a2+ab+b2)c3)=(ab)(a2bab2+ca2+cab+cb2c3)=(ab)(ca2a2b+cabab2c3+cb2)=(ab)(a2(cb)+ab(cb)c(c2b2))=(ab)(a2(cb)+ab(cb)c(cb)(c+b))=(ab)(cb)(a2+abc(c+b))=(ab)(cb)(a2+abc2cb)=(ab)(cb)(a2c2+abcb)=(ab)(cb)((ac)(a+c)+b(ac))=(ab)(cb)(ac)(a+c+b)=(ab)(cb)(ac)(a+b+c)

 

laugh

 Jun 5, 2020

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