### Video Transcript

In this video, we will learn how to
factor the sum and the difference of two cubes. We will begin by showing the
formulas we can use to factor or factorize the sum of two cubes and the difference
of two cubes.

A polynomial in the form 𝑎 cubed
plus 𝑏 cubed is called a sum of two cubes. Any polynomial of this form can be
factored such that 𝑎 cubed plus 𝑏 cubed is equal to 𝑎 plus 𝑏 multiplied by 𝑎
squared minus 𝑎𝑏 plus 𝑏 squared. We can prove that this is the case
by expanding the brackets or distributing the parentheses on the right-hand
side. We begin by distributing 𝑎. We multiply 𝑎 squared, negative
𝑎𝑏, and 𝑏 squared by 𝑎. This gives us 𝑎 cubed minus 𝑎
squared 𝑏 plus 𝑎𝑏 squared.

We then distribute the 𝑏. This gives us 𝑎 squared 𝑏 minus
𝑎𝑏 squared plus 𝑏 cubed. We notice that we can cancel an 𝑎
squared 𝑏 as negative 𝑎 squared 𝑏 plus 𝑎 squared 𝑏 is equal to zero. We can also cancel 𝑎𝑏
squared. This leaves us with 𝑎 cubed plus
𝑏 cubed, which is equal to the left-hand side. This proves that 𝑎 cubed plus 𝑏
cubed is equal to 𝑎 plus 𝑏 multiplied by 𝑎 squared minus 𝑎𝑏 plus 𝑏
squared.

Now let’s consider a polynomial in
the form 𝑎 cubed minus 𝑏 cubed. This is called the difference of
two cubes. 𝑎 cubed minus 𝑏 cubed can be
factored into the following form: 𝑎 minus 𝑏 multiplied by 𝑎 squared plus 𝑎𝑏
plus 𝑏 squared. Once again, we can prove this by
distributing the parentheses. Multiplying 𝑎 by 𝑎 squared plus
𝑎𝑏 plus 𝑏 squared gives us 𝑎 cubed plus 𝑎 squared 𝑏 plus 𝑎𝑏 squared. Multiplying the three terms in the
second bracket by negative 𝑏 gives us negative 𝑎 squared 𝑏 minus 𝑎𝑏 squared
minus 𝑏 cubed.

Once again, the 𝑎 squared 𝑏 and
𝑎𝑏 squared terms cancel, leaving us with 𝑎 cubed minus 𝑏 cubed. 𝑎 cubed minus 𝑏 cubed is equal to
𝑎 minus 𝑏 multiplied by 𝑎 squared plus 𝑎𝑏 plus 𝑏 squared. In all the questions that follow in
this video, we will need to use one of these two formulae. In each case, we will rewrite the
appropriate formula so that by the end of this lesson, we will hopefully have
learned them both.

Given that 𝑥 cubed minus 512 is
equal to 𝑥 minus eight multiplied by 𝑥 squared plus 𝑘 plus 64, find an expression
for 𝑘.

Any expression in the form 𝑎 cubed
minus 𝑏 cubed is known as the difference of two cubes. We know that this can be factored
into the form 𝑎 minus 𝑏 multiplied by 𝑎 squared plus 𝑎𝑏 plus 𝑏 squared. In this question, our value of 𝑎
cubed is 𝑥 cubed and our value of 𝑏 cubed is 512. If 𝑎 cubed is equal to 𝑥 cubed,
we know that 𝑎 is equal to 𝑥 as we can cube root both sides of the equation. If 𝑏 cubed equals 512, then 𝑏 is
equal to eight. We know this as eight cubed is
equal to 512 which means that the cube root of 512 is eight.

When factoring 𝑎 cubed minus 𝑏
cubed, the first set of parentheses contained 𝑎 minus 𝑏. This means that in our example, we
will have 𝑥 minus eight. The first term in the second set of
parentheses will be 𝑥 squared. The second term is 𝑎 multiplied by
𝑏, which is eight 𝑥. The final term is 𝑏 squared, which
in our case is eight squared, which equals 64. We are asked to find an expression
for 𝑘. This is equal to eight 𝑥. It is the value of 𝑎 multiplied by
𝑏.

In our next question, we will need
to factorize the sum of two cubes fully.

The expression 𝑥 cubed plus 27 has
two factors. One factor is 𝑥 plus three. What is the other factor?

We recall that any expression
written in the form 𝑎 cubed plus 𝑏 cubed is known as the sum of two cubes. This can be factored into two sets
of parentheses, 𝑎 plus 𝑏 multiplied by 𝑎 squared minus 𝑎𝑏 plus 𝑏 squared. In this question, 𝑎 cubed is equal
to 𝑥 cubed and 𝑏 cubed is equal to 27. We can work out the values of 𝑎
and 𝑏 by cube rooting both sides of these equations. This gives us values of 𝑎 and 𝑏
of 𝑥 and three, respectively. We can now use this information to
factor 𝑥 cubed plus 27 into two sets of parentheses.

The first bracket is 𝑎 plus
𝑏. In our question, this is 𝑥 plus
three. We were already told in the
question that this was one of the factors. Our second set of parentheses is
equal to 𝑎 squared minus 𝑎𝑏 plus 𝑏 squared. 𝑎 squared is equal to 𝑥
squared. 𝑎 multiplied by 𝑏 is equal to
three 𝑥, so our second term is negative three 𝑥. 𝑏 squared is equal to nine as
three multiplied by three is nine. The expression 𝑥 cubed plus 27 can
be factored into the form 𝑥 plus three multiplied by 𝑥 squared minus three 𝑥 plus
nine. This means that the correct answer
to the question is 𝑥 squared minus three 𝑥 plus nine. This is the other factor of 𝑥
cubed plus 27.

We could check this answer by
distributing our parentheses. We could multiply 𝑥 squared minus
three 𝑥 plus nine by 𝑥 and then multiply 𝑥 squared minus three 𝑥 plus nine by
three. When we do this, all the terms
would cancel with the exception of 𝑥 cubed plus 27.

Our next question is a more
complicated problem as we need to take out the highest common factor first.

Factorize fully 1,000𝑥 cubed minus
125.

At first glance, it appears that
this expression is written in the form 𝑎 cubed minus 𝑏 cubed, which is the
difference of two cubes. We know that any expression of this
type can be factored as shown into two sets of parentheses, 𝑎 minus 𝑏 multiplied
by 𝑎 squared plus 𝑎𝑏 plus 𝑏 squared. However, when we consider the
numbers 1,000 and 125, we notice they have a highest common factor that is greater
than one. In fact, 1,000 and 125 are both
divisible by 125. This means that we can begin this
question by factoring out 125. 1,000 divided by 125 is equal to
eight. This means that 125 multiplied by
eight 𝑥 cubed is equal to 1,000𝑥 cubed. As 125 divided by 125 is equal to
one, 1,000𝑥 cubed minus 125 is equal to 125 multiplied by eight 𝑥 cubed minus
one.

The expression eight 𝑥 cubed minus
one is still in the form 𝑎 cubed minus 𝑏 cubed, which means that this can be
factored further. 𝑎 cubed is equal to eight 𝑥
cubed, and 𝑏 cubed is equal to one. We can then cube root both sides of
these equations to calculate the values of 𝑎 and 𝑏. The cube root of eight is equal to
two. Therefore, 𝑎 is equal to two
𝑥. The cube root of one is one, so 𝑏
is equal to one. We can now factorize eight 𝑥 cubed
minus one into our two parentheses.

𝑎 minus 𝑏 is equal to two 𝑥
minus one. 𝑎 squared is equal to four 𝑥
squared as two 𝑥 multiplied by two 𝑥 is four 𝑥 squared. Multiplying our values of 𝑎 and 𝑏
gives us two 𝑥. Finally, 𝑏 squared is equal to
one. Eight 𝑥 cubed minus one is,
therefore, equal to two 𝑥 minus one multiplied by four 𝑥 squared plus two 𝑥 plus
one. We can, therefore, conclude that
the fully-factored form of 1,000 𝑥 cubed minus 125 is 125 multiplied by two 𝑥
minus one multiplied by four 𝑥 squared plus two 𝑥 plus one.

In our next question, we need to
work out the sum of two cubes by distributing parentheses.

Complete the following: Blank is
equal to 𝑦 plus 15𝑥 multiplied by 𝑦 squared minus 15𝑦𝑥 plus 225𝑥 squared.

Our first thought in this question
might be to try and distribute the parentheses, to multiply 𝑦 squared minus 15𝑦𝑥
plus 225𝑥 squared firstly by 𝑦 and then by 15𝑥. However, we might notice that our
expression is written in the form 𝑎 plus 𝑏 multiplied by 𝑎 squared minus 𝑎𝑏
plus 𝑏 squared. This is the factored form of the
expression 𝑎 cubed plus 𝑏 cubed. This is known as the sum of two
cubes. Our value of 𝑎 is 𝑦, and our
value of 𝑏 is 15𝑥.

We can work out the value of 𝑎
cubed and 𝑏 cubed by cubing both sides of each of these equations. Our first equation gives us 𝑎
cubed is equal to 𝑦 cubed. Cubing both sides of our second
equation gives us 𝑏 cubed is equal to 3,375𝑥 cubed. This is because 15 multiplied by 15
multiplied by 15 is 3,375. The missing term is, therefore, 𝑦
cubed plus 3,375𝑥 cubed as this is equal to 𝑦 plus 15𝑥 multiplied by 𝑦 squared
minus 15𝑦𝑥 plus 225𝑥 squared.

We would’ve got the same answer had
we distributed the two sets of parentheses. All the terms would’ve canceled
with the exception of 𝑦 multiplied by 𝑦 squared, which is 𝑦 cubed, and 15𝑥
multiplied by 225𝑥 squared, which is 3,375𝑥 cubed.

In our final question, we have a
more complicated initial expression.

Factorize fully 𝑥 minus six 𝑦 all
cubed minus 216𝑦 cubed.

Whilst it might not be immediately
obvious, this expression is written in the form 𝑎 cubed minus 𝑏 cubed. It is the difference of two
cubes. We know that the factorization of
𝑎 cubed minus 𝑏 cubed is equal to 𝑎 minus 𝑏 multiplied by 𝑎 squared plus 𝑎𝑏
plus 𝑏 squared. The first term in our expression is
𝑥 minus six 𝑦 all cubed. This means that 𝑎 cubed is equal
to this. As cube rooting is the opposite of
cubing, we can cube root both sides of this equation, giving us a value of 𝑎 equal
to 𝑥 minus six 𝑦. Our second term is 216𝑦 cubed;
therefore, 𝑏 cubed is equal to this. Once again, we can cube root both
sides of this equation, giving us 𝑏 is equal to six 𝑦 as the cube root of 216 is
six.

We can now substitute our values of
𝑎 and 𝑏 into the right-hand side of the formula. We begin with 𝑎 minus 𝑏. This is equal to 𝑥 minus six 𝑦
minus six 𝑦. This simplifies to 𝑥 minus
12𝑦. 𝑎 squared will be equal to 𝑥
minus six 𝑦 all squared. This is equal to 𝑥 minus six 𝑦
multiplied by 𝑥 minus six 𝑦. Distributing our parentheses here
gives us 𝑥 squared minus six 𝑦𝑥 minus six 𝑦𝑥 plus 36𝑦 squared. This can be simplified to 𝑥
squared minus 12𝑦𝑥 plus 36𝑦 squared. 𝑎𝑏 is equal to 𝑥 minus six 𝑦
multiplied by six 𝑦. Distributing the parentheses here
gives us six 𝑦𝑥 minus 36𝑦 squared.

Finally, 𝑏 squared is equal to six
𝑦 all squared. This is equal to 36𝑦 squared. Substituting in the replacement for
these three terms gives us 𝑥 squared minus 12𝑦𝑥 plus 36𝑦 squared plus six 𝑦𝑥
minus 36𝑦 squared plus 36𝑦 squared. This can be simplified to 𝑥
squared minus six 𝑦𝑥 plus 36𝑦 squared. Six 𝑦𝑥 is the same as six
𝑥𝑦. Therefore, the fully factorized
form is 𝑥 minus 12𝑦 multiplied by 𝑥 squared minus six 𝑥𝑦 plus 36𝑦 squared.

We will now summarize the key
points from this video. We can factorize or factor the sum
of two cubes and difference of two cubes using the following formulae. 𝑎 cubed plus 𝑏 cubed is equal to
𝑎 plus 𝑏 multiplied by 𝑎 squared minus 𝑎𝑏 plus 𝑏 squared. 𝑎 cubed minus 𝑏 cubed on the
other hand is equal to 𝑎 minus 𝑏 multiplied by 𝑎 squared plus 𝑎𝑏 plus 𝑏
squared. Before using either of the two
formulae, it is important that we factor out the highest common factor of the two
terms first.