HELPMEEEEEEEEEEEEE

You can get your answer here. I wouldn't recommend this website to use everytime you have a question though.

Hope it helps!

You already posted this question here. If I did not answer them correctly, wait for someone else to answer it.

Thanks!

This question has been answered here by ElectricPavlov and CPhill.

Please check if your question has been answered before posting it.

Thanks!

Hope it helps!

The way I would approach this is by factoring by grouping.

ab^3-a^3b= (ab)(b^2-a^2)

bc^3-b^3c= (bc)(c^2-b^2)

ca^3-c^3a= (ca)(a^2-c^2)

Using difference of squares...

ab^3-a^3b= (ab)(b^2-a^2)= (ab)(b+a)(b-a)

bc^3-b^3c= (bc)(c^2-b^2)= (bc)(c+b)(c-b)

ca^3-c^3a= (ca)(a^2-c^2)= (ca)(a+c)(a-c)

Putting the whole thing together, you get (ab)(b+a)(b-a)+ (bc)(c+b)(c-b)+ (ca)(a+c)(a-c)​ which is completely factored.

Hope it helps!