### Video Transcript

Prove algebraically that four lots
of π plus one all squared minus π minus two all squared is always a multiple of
three.

First, letβs note that any multiple
of three can be written as three π, where π is an integer. For example, three equals three
times one. Six equals three times two. And 18 equals three times six. And we could do this for any
multiple of three. We will use this rule for writing
multiples of three in order to help us answer the question.

Letβs now look at the expression
given in the question. We will start by expanding π plus
one squared and π minus two squared, using the FOIL method, starting with π plus
one squared, which we can also write as π plus one times π plus one.

And now we will use the FOIL
method. So we start by multiplying the
first term in each bracket. Thatβs π times π, giving us π
squared. Then we multiply the outside
terms. So thatβs π times one, giving us
plus π. Then we multiply the inside
terms. Thatβs one times π, giving us
another plus π. And then, finally, we multiply the
last term in each bracket. So thatβs one times one, giving us
plus one.

Now we notice that we can combine
the π plus π to give us two π. And we obtain that π plus one all
squared is equal to π squared plus two π plus one. Now we will expand π minus two
squared, which can be written as π minus two times π minus two. And now we will use the FOIL
method.

So we start by multiplying the
first term in each bracket. This is π times π, giving us π
squared. Next, we multiply the outside term
in each bracket. So this is π timesed by minus two,
giving us minus two π. Next, we multiply the inside
terms. So thatβs minus two times π,
giving us another minus two π. And then, finally, we multiply the
last term in each bracket. So this is minus two times minus
two. And since a negative times a
negative gives us a positive, this is equal to plus four.

Now we notice that we have minus
two π minus two π, giving us minus four π. And so we get that π minus two all
squared is equal to π squared minus four π plus four. Now we can substitute the expanded
forms of π plus one squared and π minus two squared back into the expression in
the question. This gives us that four lots of π
plus one squared minus π minus two squared is equal to four timesed by π squared
plus two π plus one minus π squared minus four π plus four.

Now we simply multiply every term
in the left-hand bracket by four and every term in the right-hand bracket by
negative one. This gives us four π squared plus
eight π plus four minus π squared plus four π minus four.

Now we can combine the terms with
the same powers of π. For the π squared terms, we have a
four π squared minus π squared. And this leaves us with three π
squared. For the π terms, we have eight π
plus four π, giving us plus 12π. And for the units, we have plus
four minus four, giving us plus zero. However, this has no effect here,
so we can ignore it.

We are left with three π squared
plus 12π. And we notice that each term here
is a multiple of three. So we can factorise out three. So we are left with three lots of
π squared plus four π. And now since π is an integer,
this means that π squared plus four π will also be an integer. And since we are multiplying an
integer by three, that means that this is always a multiple of three, using the rule
which we stated at the start. And so we have shown that four lots
of π plus one all squared minus π minus two all squared is always a multiple of
three.