### Video Transcript

Simplify five minus π to the 83rd
power multiplied by five minus π to the 69th power multiplied by five minus π to
the 61st power.

In this question, weβre asked to
simplify the product of three complex numbers. However, we can see that all three
of our products share something interesting. They contain π raised to a very
high exponent. We mightβve been tempted to just
start distributing our parentheses straight away. However, in this case, weβre going
to simplify whatβs inside our parentheses first.

To do this, letβs start by
recalling that we define π to be the square root of negative one. And this gives us a very useful
property. We can ask the question, what
happens if we square both sides of this equation? We get that π squared is equal to
negative one. And we can immediately use this to
simplify all three of our factors.

Letβs start with negative π to the
83rd power. By writing this product out in full
and simplifying, all by using our laws of exponents, we can see this is equal to
negative one multiplied by π to the 82nd power multiplied by π. But π to the 82nd power can be
written in terms of π squared. Either by writing the product out
in full and simplifying or by using our laws of exponents, we can see π to the 82nd
power is π squared all raised to the 41st power. But we know π squared is equal to
negative one. So this simplifies to give us
negative one times negative one to the 41st power multiplied by π.

And now we can evaluate this
expression by using the fact that negative one raised to an even power is equal to
one and negative one raised to an odd power is equal to negative one. This means negative one raised to
the power 41 is negative one. And then we multiply this by
negative one to just get one. So this entire expression just
simplified to give us π. So letβs keep note of what weβve
shown. So far, weβve shown that negative
one times π to the 83rd power is just equal to π.

We can now do the same for our
second term. This time, weβre going to want to
simplify negative one times π to the 69th power. Weβll start by rewriting this as
negative one times π to the 68th power multiplied by π. Then, once again, we use the fact
that π to the 68th power is equal to π squared all raised to the 34th power to
rewrite our expression. Then all we use is the fact that π
squared is equal to negative one. This gives us negative one times
negative one to the 34th power multiplied by π. And because negative one raised to
the 34th power is negative one raised to an even power, this is just equal to
one. So this all simplifies to just give
us negative π.

Once again, letβs keep note of
this. Negative one times π to the 69th
power is equal to negative π. So letβs clear some space and do
the same for our last factor. This time, we want to simplify
negative one times π to the 61st power. Weβll start by writing this as
negative one times π to the 60th power multiplied by π. Then we rewrite π to the 60th
power as π squared all raised to the 30th power. Then we just use the fact that π
squared is equal to negative one to write this as negative one times negative one to
the 30th power multiplied by π. And of course negative one raised
to the 30th power is just equal to one. So this all simplifies to give us
negative π.

And letβs keep note of this last
simplification because weβll need it for later. So letβs clear some space, so weβre
now ready to start simplifying our expression. First, weβll start with our
expression in full. Thatβs five minus π to the 83rd
power multiplied by five minus π to the 69th power times five minus π to the 61st
power.

Next, weβre going to use our three
simplifications. This lets us write our first factor
as five plus π, our second factor as five minus π, and our third factor also as
five minus π. But remember, the question wants us
to simplify this expression. And we could simplify this by
distributing all of our parentheses.

Now, we could do this factor by
factor by using the FOIL method. However, thereβs an easier
method. We can see that the first two
factors in this expression is actually a factoring of a difference between
squares. Itβs of the form π₯ plus π¦
multiplied by π₯ minus π¦. And we know this is equal to π₯
squared minus π¦ squared. So instead of multiplying this out
in full, we already know their products will be five squared minus π squared. This gives us five squared minus π
squared multiplied by five minus π.

Next, we can simplify this by
remembering that π squared is equal to negative one. This gives us 25 minus negative one
all multiplied by five minus π. And of course subtracting negative
one is the same as adding one. And 25 plus one is equal to 26. So we get 26 multiplied by five
minus π. And finally, weβll distribute 26
over our parentheses. And since 26 multiplied by five is
130, this gives us our final answer of 130 minus 26π.

Therefore, we were able to simplify
the expression five minus π to the 83rd power times five minus π to the 69th power
multiplied by five minus π to the 61st power. We were able to show itβs equal to
130 minus 26π.