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Current time:0:00Total duration:4:12

CCSS Math: HSA.SSE.A.2

We're asked to factor 40c
to the third power minus 5d to the third power. So the first thing that
might jump out at you is that 5 is a factor
of both of these terms. I could rewrite this as 5
times 8c to the third power minus 5 times d to
the third power. And so you could actually
factor out a 5 here, so factor out a 5. And so if you
factor out a 5, you get 5 times 8c to
the third power minus d to the third power. So as you see,
factoring, it really is just undistributing
the 5, reversing the distributive property. And when you write
it like this, it might jump out at you
that 8 is a perfect cube. It's 2 to the third power.
c to the third power is obviously c to
the third power. And then you have d
to the third power. So this right here is
a difference of cubes. And actually, let me
write that explicitly because 8 is the same thing
as 2 to the third power. So you can write this as-- let
me write the 5 out front-- 5 times-- this term
right over here can be rewritten as
2c to the third power because it's 2 to the
third power times c to the third power-- 8c to
the third power-- And then, minus d to the third power. And so this gives us, right over
here, a difference of cubes. And you can actually factor
a difference of cubes. And you may or may
not know the pattern. So if I have a to the
third minus b to the third, this can be factored
as a minus b times a squared plus
ab plus b squared. And if you don't believe me, I
encourage you to multiply this out, and you will get a to the
third minus b to the third. You get a bunch of
terms that cancel out, so you're only left
with two terms. And even though it's
not applicable here, it's also good to know
that the sum of cubes is also factorable. It's factorable as a plus b
times a squared minus ab plus b squared. So once again, I won't go
through the time of multiplying this out, but I
encourage you to do so. It just takes a little bit
of polynomial multiplication. And you'll be able
to prove to yourself that this is indeed the case. Now, assuming that
this is the case, we can just do a little
bit of pattern matching. Because in this case, our
a is 2c, and our b is d. So let me write this. a is equal
to 2c, and our b is equal to d. We have minus b to the third
and minus d to the third, so b and d must
be the same thing. So this part inside
must factor out to-- let me write my
5, open parentheses. Let me give myself some space. So it's going to factor
out into a minus b. So a is 2c minus b, which is d. So it factors out as the
difference of the two things that I'm taking the cube of. 2c minus d times-- and now, I
have a squared is 2c squared. 2c squared is the same
thing as 4c squared. Let me make that. a squared is
equal to 2c-- the whole thing squared, which is
equal to 4c squared. So it's 4c squared
plus a times b. So that's going to be 2c
times d, so plus 2c times d. And then finally plus b squared,
and in our case, b is d. So you get plus d squared. And you're done. We have factored it out. And actually, you could get
rid of one set of parentheses. This can be factored as 5
times 2c minus d times 4c squared plus 2cd plus d squared. And we are done.