### Video Transcript

Simplify one minus π cubed minus
one plus π times one minus π plus one plus π cubed.

To simply this expression, weβll
begin by distributing each set of parentheses. Letβs begin by distributing one
minus π cubed. Now, we could write this as one
minus π times one minus π times one minus π and then distribute the first pair of
parentheses and then multiply all of that by one minus π.

The binomial theorem tells us that
for positive integer values of π, π plus π to the πth power is the sum from π
equals zero to π of π choose π times π to the πth power times π to the power
of π minus π. So, in the distribution of one
minus π cubed, the first term is one cubed. Then, the second term is three
choose one times one squared times negative π to the power of one.

The third term is three choose two
times one times negative π squared. And then, the third term is
negative π cubed. Letβs simplify this a little. One cubed is one, and three choose
one is three. So, our second term becomes three
times one times negative π, which is negative three π. Our third term is three choose two,
which is three times π squared. But of course, we know that π
squared is negative one. So, this simply becomes negative
three.

And our third term is negative π
cubed. But since we can write π cubed as
π squared times π, and we know π squared is negative one, this becomes negative
negative one times π, which is simply π. And so, we simplify fully. And we find that one minus π cubed
is negative two minus two π.

Weβll now repeat this process for
our second term. Thatβs one plus π times one minus
π. Now, we could distribute the
parentheses using something like the FOIL method. Or we can recall that the product
of a complex number and its conjugate is the sum of the squares of the real and
imaginary parts of that complex number. And remember, the complex conjugate
of one plus π is found by changing the sign of the imaginary part, so itβs one
minus π. Its real part is one, and its
imaginary part is the coefficient of π, so itβs also one. So, one plus π times one minus π
is one squared plus one squared, which is simply two.

Letβs move on to the third pair of
parentheses. Weβve got one plus π cubed. Once again, we use the binomial
theorem. And we get one cubed plus three
choose one times one squared times π plus three choose two times one times π
squared plus π cubed. We simplify a little, and we get
one plus three π plus three π squared plus π cubed. And then, we use again the fact
that π squared is negative one. So, our third term becomes negative
three. And our fourth term becomes
negative π.

Simplifying this time, and we find
that one plus π cubed is negative two plus two π. We can now replace each parentheses
in our original expression with these. We get negative two minus two π
minus two plus negative two plus two π. Then, we see that negative two π
plus two π is zero. So, this becomes negative two minus
two minus two, which is negative six. And so, when we simplify one minus
π cubed minus one plus π times one minus π plus one plus π cubed, we get
negative six.