Factoring the expression

[thecat]

How can I factor the expression below only using basic factoring techniques? thank you in advance.

\(\displaystyle a^2b+a^2c+ab^2+b^2c+ac^2+bc^2−a^3−b^3−c^3−2abc \)

 

[Deleted member 4993]

How can I factor the expression below only using basic factoring techniques? thank you in advance.

\(\displaystyle a^2b+a^2c+ab^2+b^2c+ac^2+bc^2−a^3−b^3−c^3−2abc \)

Please share your work/thoughts about this assignment.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

 

[Dr.Peterson]

I gave it a try and it worked out. I just arranged the terms in descending order by power of a, factored the last term by grouping, and then factored the whole thing by grouping. It took a little skill, but no advanced knowledge.

 

[thecat]

I gave it a try and it worked out. I just arranged the terms in descending order by power of a, factored the last term by grouping, and then factored the whole thing by grouping. It took a little skill, but no advanced knowledge.

Can you show?

 

[thecat]

Please share your work/thoughts about this assignment.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

I kept trying but couldn't get anything productive

 

[firemath]

First, find like terms, such as all with a or all with b.Then factor out those common variables in descending order of the variable.

 

[Steven G]

Please try what Dr Peterson tried and show us how far you get. We want to help you by not just giving you the answer.

 

[Dr.Peterson]

I kept trying but couldn't get anything productive

Then show something unproductive. If even your first step is valid, we will have something to start with. If you make a mistake, that will give us more to talk about.

 

[thecat]

Then show something unproductive. If even your first step is valid, we will have something to start with. If you make a mistake, that will give us more to talk about.

I looked again later and made it here

\(\displaystyle -a^3 + a^2b + a^2c + abc + b^2c + bc^2 -b^3 - c^3 -3abc \)

\(\displaystyle \therefore a^2(-a+b+c) + a(b^2 + c^2 + bc) +bc(b+c) -(b+c)(b^2+bc+c^2 - 2bc) -3abc \)

\(\displaystyle \therefore a^2(-a+b+c) + a(b^2+bc+c^2) +bc(b+c) -(b+c)(b^2+bc+c^2) +(b+c)2bc -3abc\)

\(\displaystyle \therefore a^2(-a+b+c) -(b^2+bc+c^2)(-a+b+c) +bc(b+c-a) +2bc(b+c-a)\)

\(\displaystyle \therefore (-a+b+c)(a^2-b^2-bc-c^2 +bc+2bc)\)

\(\displaystyle \therefore (-a+b+c)(a^2 -b^2-c^2 + 2bc)\)

\(\displaystyle \therefore (-a+b+c)(a^2-[b-c]^2) \)

\(\displaystyle \therefore(-a+b+c)(a-b+c)(a+b-c)\).

 

[thecat]

Then show something unproductive. If even your first step is valid, we will have something to start with. If you make a mistake, that will give us more to talk about.

I looked again later and made it here
\(\displaystyle -a^3 + a^2b + a^2c + abc + b^2c + bc^2 -b^3 - c^3 -3abc \)

\(\displaystyle \therefore a^2(-a+b+c) + a(b^2 + c^2 + bc) +bc(b+c) -(b+c)(b^2+bc+c^2 - 2bc) -3abc \)

\(\displaystyle \therefore a^2(-a+b+c) + a(b^2+bc+c^2) +bc(b+c) -(b+c)(b^2+bc+c^2) +(b+c)2bc -3abc\)

\(\displaystyle \therefore a^2(-a+b+c) -(b^2+bc+c^2)(-a+b+c) +bc(b+c-a) +2bc(b+c-a)\)

\(\displaystyle \therefore (-a+b+c)(a^2-b^2-bc-c^2 +bc+2bc)\)

\(\displaystyle \therefore (-a+b+c)(a^2 -b^2-c^2 + 2bc)\)

\(\displaystyle \therefore (-a+b+c)(a^2-[b-c]^2) \)

\(\displaystyle \therefore(-a+b+c)(a-b+c)(a+b-c)\).

 

[Dr.Peterson]

Your method is a little different from mine. Here is what I did:

[MATH]a^2b+a^2c+ab^2+b^2c+ac^2+bc^2−a^3−b^3−c^3−2abc[/MATH][MATH]−a^3 + a^2b+a^2c+ab^2−2abc+ac^2−b^3+b^2c+bc^2−c^3[/MATH] (descending order)
[MATH]−[a^3 - a^2b-a^2c-ab^2+2abc-ac^2+b^3-b^2c-bc^2+c^3][/MATH] (factor out negative)
[MATH]−[a^3 - (b+c)a^2 - (b^2-2bc+c^2)a +(b^2 - c^2)(b-c)][/MATH] (factor out powers of a)
[MATH]−[a^3 - (b+c)a^2 - (b-c)^2a +(b-c)^2(b+c)][/MATH] (factor each coefficient)
[MATH]−[a^2(a - (b+c)) - (b-c)^2(a -(b+c))][/MATH] (factor by grouping)
[MATH]−[(a^2- (b-c)^2)(a -(b+c))][/MATH][MATH]−(a+b-c)(a-b+c)(a -b-c)[/MATH] (factor difference of squares)

 

[Steven G]

I looked again later and made it here
\(\displaystyle -a^3 + a^2b + a^2c + abc + b^2c + bc^2 -b^3 - c^3 -3abc \)

\(\displaystyle \therefore a^2(-a+b+c) + a(b^2 + c^2 + bc) +bc(b+c) -(b+c)(b^2+bc+c^2 - 2bc) -3abc \)

\(\displaystyle \therefore a^2(-a+b+c) + a(b^2+bc+c^2) +bc(b+c) -(b+c)(b^2+bc+c^2) +(b+c)2bc -3abc\)

\(\displaystyle \therefore a^2(-a+b+c) -(b^2+bc+c^2)(-a+b+c) +bc(b+c-a) +2bc(b+c-a)\)

\(\displaystyle \therefore (-a+b+c)(a^2-b^2-bc-c^2 +bc+2bc)\)

\(\displaystyle \therefore (-a+b+c)(a^2 -b^2-c^2 + 2bc)\)

\(\displaystyle \therefore (-a+b+c)(a^2-[b-c]^2) \)

\(\displaystyle \therefore(-a+b+c)(a-b+c)(a+b-c)\).

1st of all please use equal signs and not the \(\displaystyle \therefore \) symbols. At best what you have is that the last line is a result of the 1st line. That does not imply that they are equal.
I need some help understanding you 2nd line of your computations. One term you have is a(b^2 + c^2 + bc) which equals (ab^2 + ac^2 + abc).
Now the 1st equation does not have a ab^2 term or a ac^2 term. This doesn't necessarily mean much but I do not see that you are subtracting ab^2 and ac^2 in that 2nd line. So how can the two lines be equal?

 

[pka]

a2b+a2c+ab2+b2c+ac2+bc2−a3−b3−c3−2abc

Look at this

 

[thecat]

1st of all please use equal signs and not the \(\displaystyle \therefore \) symbols. At best what you have is that the last line is a result of the 1st line. That does not imply that they are equal.
I need some help understanding you 2nd line of your computations. One term you have is a(b^2 + c^2 + bc) which equals (ab^2 + ac^2 + abc).
Now the 1st equation does not have a ab^2 term or a ac^2 term. This doesn't necessarily mean much but I do not see that you are subtracting ab^2 and ac^2 in that 2nd line. So how can the two lines be equal?

1st of all please use equal signs and not the \(\displaystyle \therefore \) symbols. At best what you have is that the last line is a result of the 1st line. That does not imply that they are equal.
I need some help understanding you 2nd line of your computations. One term you have is a(b^2 + c^2 + bc) which equals (ab^2 + ac^2 + abc).
Now the 1st equation does not have a ab^2 term or a ac^2 term. This doesn't necessarily mean much but I do not see that you are subtracting ab^2 and ac^2 in that 2nd line. So how can the two lines be equal?

noticed that I forgot to rewrite the factors \(\displaystyle ab^2, ac^2 \) in the first line. If you look at the question, both are there. Thanks for the tip, I'm going to use "equal" instead of "therefore"