### Video Transcript

Simplify the function π of π₯ is
equal to π₯ squared minus 81 over π₯ cubed plus 729 and find its domain.

In order to simplify the function
π of π₯, weβll need to factorize the numerator and denominator. In order to do this, we need to
recognize that on the numerator, we have the difference of two squares. This tells us that any quadratic of
the form π₯ squared minus π squared is equal to π₯ plus π multiplied by π₯ minus
π. We know that nine squared is equal
to 81. This means that π₯ squared minus 81
is equal to π₯ plus nine multiplied by π₯ minus nine. The denominator π₯ cubed plus 729
is written in the form π₯ cubed plus π cubed.

Once again, π is equal to nine as
nine cubed is 729. π₯ cubed plus π cubed is equal to
π₯ plus π multiplied by π₯ squared minus ππ₯ plus π squared. This means that π₯ cubed plus 729
is equal to π₯ plus nine multiplied by π₯ squared minus nine π₯ plus 81. We can divide the numerator and
denominator by π₯ plus nine, so these terms cancel. The simplified version of π of π₯
is therefore equal to π₯ minus nine over π₯ squared minus nine π₯ plus 81.

The second part of our question
asks us to find the domain. The domain of a function is the set
of π₯-values we can substitute or input into π of π₯. As the denominator of a fraction
cannot be equal to zero, the domain will be all real values apart from those which
make the denominator zero. To calculate these values, we set
π₯ cubed plus 729 equal to zero. We already know this is equal to π₯
plus nine multiplied by π₯ squared minus nine π₯ plus 81. Letβs consider the linear term π₯
plus nine first. When π₯ plus nine is equal to zero,
π₯ is equal to negative nine. Therefore, negative nine cannot be
in the domain of π of π₯. The quadratic π₯ squared minus nine
π₯ plus 81 cannot be factored or factorized.

We can actually go one stage
further here as it actually has no real solutions. We know that any quadratic of the
form ππ₯ squared plus ππ₯ plus π has no real solutions when its discriminant, π
squared minus four ππ, is less than zero. In this case, we have π is equal
to one, the coefficient of π₯ squared, π is equal to negative nine, the coefficient
of π₯, and π is 81. We need to calculate negative nine
squared minus four multiplied by one multiplied by 81. This is equal to negative 243. As this is less than naught, the
quadratic π₯ squared minus nine π₯ plus 81 equals zero has no solutions.

This means that the only value such
that π₯ cubed plus 729 equals zero is π₯ is equal to negative nine. The domain of π of π₯ is therefore
equal to all real values apart from negative nine. We can write this in set notation
as shown.