Question

Can someone help me factor this equation? I just need you to explain some steps. Thanks.

The equation is

m^3-3m^2a+3ma^2-a^3

Answer 1

\[m^3-3m^2a+3ma^2-a^3\]

Answer 2

factor by grouping

Answer 3

I have factored by grouping but i got confused on something. I'll show my work and explain what i'm confused on

Answer 4

sounds good

Answer 5

It just so happens that the expression you've shared is the cube of a binomial. Know where to find a formula for factoring that? Factoring by grouping should also work. In what way do you feel stuck?

Answer 6

\[(m^3-a^3) + (-3m^2a+3ma^2)\]
\[(m-a) (m^2+ma+a^2)-3ma(m-a)\]

Answer 7

i have gotten stuck after this step

Answer 8

i believe that you square these two to get
\[(m-a)^2 (m^2+ma+a^2)-3ma\]

Answer 9

\(\bf {\color{blue}{ (m-a) }} (m^2+ma+a^2)-3ma{\color{blue}{(m-a) }}\) notice any common factor?

Answer 10

this is where my confusion is. After this step, I'm unsure of what to do next.

Answer 11

can someone explain what i do after getting \[(m-a)^2 (m^2+ma+a^2) -3ma\]

Answer 12

I'm unsure of what to do with the -3ma

Answer 13

well... hold the mayo

Answer 14

but thus far, you're correct, one sec

Answer 15

\({\color{blue}{ (m-a) }} (m^2+ma+a^2)-3ma{\color{blue}{(m-a) }}
\\ \quad \\
(m-a)[(m^2+ma+a^2)-3ma]
\\ \quad \\
(m-a)[m^2{\color{brown}{ -3ma+ma}}+a^2]
\\ \quad \\
(m-a)[m^2-2ma+a^2]
\\ \quad \\
\textit{now, recall what a perfect square trinomial is}
\\ \quad \\
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad

% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\)

and notice, that \(\bf (m-a)[{\color{purple}{ m^2-2ma+a^2 }} ]\impliedby {\color{purple}{ \textit{perfect square trinomial} }}
\)

Answer 16

Question, where did the second (m-a) go in the equation though? I understand most of this but I don't understand why the (m-a) isn't squared.

Answer 17

notice the 3rd line, the "brown" part, the addition

Answer 18

m-a isn't squared? hmmm the first one? wasn't that the common factor you took from the second group? :)

Answer 19

No, that I understand. The -3ma +ma would equal to -2ma because of the 1 in front of the ma. But what I'm asking is why isn't the (m-a) and the (m-a) in -3ma together to make (m-a)^2

Answer 20

wait...now I'm a bit confused

Answer 21

ohhhh why would they combine into the binomial at the 2nd power? well
I gather you do not recall your "perfect square trinomial" then =)
but usually comes out of FOIL, if you do say (m-a)(m-a), or \((m-a)^2\)
you'll noitice you'll end up with \(m^2-2ma-a^2\)

Answer 22

wouldn't you get \[(m-a)^2 (m^2-2ma+a^2)\]

Answer 23

oh wait so the answer would be (m-a)^3

Answer 24

because it's a perfect square trinomial oh

Answer 25

ahemm nope
because the (m-a) on the left-side,was just the common factor of the second group
and yes, \((m-a)(m-a)^2\implies (m-a)^3\)

Answer 26

but i usually thought that you had to break apart the trinomial
\[(m^2-2ma+a^2) = (m-ma-a) (m+ma-a)\]

Answer 27

then you get (m-a) (m-a)

Answer 28

that's why i assumed it would be \[(m-a)^4\] because of the first (m-a)^2 and the two new (m-a)

Answer 29

ohhh ahemmm not quite
you're thinking \(\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\)

that's not true for a perfect square trinomial
\(\bf a^2-b^2 \ne (a-b)^2\)

Answer 30

wait, can you elaborate a bit more? I may have confused the two now. My teacher taught all the factoring at once and I may have mixed up the rules.

Answer 31

oh wait

Answer 32

so would that mean that (m-a)^2 would equal to (m-a)... no i think i confused myself

Answer 33

well... hmm what part is throwing you off? \((m-a)^2 \implies (m-a)(m-a)\)

Answer 34

i just don't understand how the \[(m-a)^2\] ends up being (m-a) and why the answer is (m-a)^3 instead of (m-a)^4

Answer 35

hmm.... lemme hmm hmmm do it from the start, shouldn't take long

Answer 36

Thank you so much

Answer 37

\(\bf m^3-3m^2a+3ma^2-a^3
\\ \quad \\
(m^3-a^3)-(3m^2a-3ma^2)
\\ \quad \\
(m-a)(m^2+ma+a^2)-(3m^2a-3ma^2)
\\ \quad \\
{\color{blue}{ (m-a) }}(m^2+ma+a^2)-3ma{\color{blue}{ (m-a) }}
\\ \quad \\
{\color{blue}{ (m-a)}}[(m^2+ma+a^2)-3ma]
\\ \quad \\
{\color{blue}{ (m-a) }}[m^2{\color{brown}{ -3ma+ma}}+a^2]
\\ \quad \\
{\color{blue}{ (m-a) }}[m^2-2ma+a^2]
\\ \quad \\

{\color{blue}{ (m-a) }}[{\color{purple}{ m^2-2ma+a^2 }} ]\impliedby {\color{purple}{ \textit{perfect square trinomial} }}
\\ \quad \\
{\color{blue}{ (m-a) }}(m-a)^2\implies (m-a)^3\)

Answer 38

Once again: There is a basic formula for factoring the difference of two cubes, and you could save a lot of time by looking that up.

Look up (e. g., through Google): special products and factors

or "factoring the difference of two cubes"

Alternatively, try assuming that (a-b) is a factor and divide that into a^3-3a^2*b-3ab^2-b^3.

Answer 39

Okay i think i understand more now. thank you all so much. I found my mistake.

Answer 40

yw