### Video Transcript

Let π equal one-half times six to
the π₯ power plus six to the negative π₯ power and π equal one-half times six to
the π₯ power minus six to the negative π₯ power. Is it true that π squared minus π
squared equals one?

If we plug in what we know for π
and π, we get the expression one-half times six to the π₯ power plus six to the
negative π₯ power squared minus one-half times six to the π₯ power minus six to the
negative π₯ power squared. Because weβre going to have to be
squaring some of these terms, I notice that we have six to the π₯ power and six to
the negative π₯ power in both the π- and the π-variable.

One strategy to make our
calculations a bit simpler would be to let π equal six to the π₯ power and π equal
six to the negative π₯ power. If we do that, we end up with
one-half times π plus π squared minus one-half times π minus π squared. We distribute our square to the
one-half and the π plus π term and to the one-half and π minus π term. One-half squared is one-fourth. And we also know the form for π
plus π squared. Itβs equal to π squared plus two
ππ plus π squared. So weβll add that. Bring down the subtraction. One-half squared is one-fourth.

And we have to be careful here
because π minus π squared is not the difference of squares. π minus π squared will be equal
to π squared minus two ππ plus π squared. So we substitute π squared minus
two ππ plus π squared back into our equation. At this point, we have a one-fourth
term that we could remove. We are undistributing that
one-fourth. So weβll have one-fourth multiplied
by π squared plus two ππ plus π squared. And then weβre subtracting this
entire term, which means weβll have negative π squared. Subtracting negative two ππ means
weβll be adding two ππ, and then subtracting a positive π squared. So we have minus π squared.

Before we do anything else, we
wanna check and see if we can simplify by combining like terms. We have a positive π squared term
and a negative π squared term, which cancels out. We have a positive π squared term
and a negative π squared term, which also cancels out. And then we have two ππ plus two
ππ. We combine them by combining their
coefficients so that we have four ππ. And if four ππ is multiplied by
one-fourth, one-fourth times four equals one. And weβre left with π times
π.

Now we need to go back in and plug
in the values we know for π and π. Weβll multiply six to the π₯ power
times six to the negative π₯ power. To do that, weβll have to remember
our exponent rules when we have the same base taken to different powers being
multiplied together. π₯ to the π power times π₯ to the
π power is equal to π₯ to the π plus π power. Since both of these exponents have
a base of six, weβll combine their powers, which will be π₯ plus negative π₯. π₯ plus negative π₯ equals zero,
and six to the zero power equals one.

Now, if we go back and look at our
original question, we want to know is π squared minus π squared equal to one. We began with π squared minus π
squared, which gave us π times π. And then we found that π times π
was equal to one. And so we can say, under these
conditions, yes, π squared minus π squared is equal to one.