### Video Transcript

Is the equation π₯ cubed minus π¦
cubed equals π₯ plus π¦ times π₯ minus π¦ times π₯ plus π¦ an identity?

Remember, an identity is an
equation thatβs true for all values of the variables involved. Here, our variables are π₯ and
π¦. Itβs simply not enough to
substitute a few values of π₯ and π¦ in and check that it works for these
values. Instead, weβre going to start with
the expression on the right-hand side, π₯ plus π¦ times π₯ minus π¦ times π₯ plus
π¦, and see how we might manipulate it to look more like a standard polynomial. Weβre going to distribute the
parentheses. To distribute three pairs of
parentheses, we begin by distributing any two. So letβs multiply π₯ plus π¦ by π₯
minus π¦.

We multiply the first term in each
expression. π₯ times π₯ gives us π₯
squared. We then multiply the outer
terms. π₯ times negative π¦ is negative
π₯π¦. We multiply the inner terms. π¦ multiplied by π₯ is another
π₯π¦. And finally we multiply the last
terms, giving us negative π¦ squared. And so we see the product of π₯
plus π¦ and π₯ minus π¦ is π₯ squared minus π₯π¦ plus π₯π¦ minus π¦ squared. But negative π₯π¦ plus π₯π¦ is
zero. And so our expression simplifies to
π₯ squared minus π¦ squared times π₯ plus π¦.

Letβs distribute again. Once again, we multiply the first
term in each expression. π₯ squared times π₯ is π₯
cubed. We multiply the outer terms, giving
us π₯ squared π¦. We then multiply the inner terms,
giving us negative π₯π¦ squared. And finally, we multiply the last
terms. Negative π¦ squared times π¦ is
negative π¦ cubed. Now here we need to note that π₯
squared π¦ and π₯π¦ squared are completely different terms. In the first term, the π₯ is being
squared before multiplying by π¦. And in the second, the π¦ is being
squared and then weβre multiplying it by π₯. So we canβt simplify any
further.

And so if we compare this
expression with our original π₯ cubed minus π¦ cubed, we see that these are not
equal. Since the expressions are not
equivalent, we donβt have an identity. This isnβt going to be true for all
values of π₯ and π¦. We can confirm this by finding a
single value of π₯ and π¦ for which the original equation doesnβt hold. Letβs try π₯ is equal to two and π¦
is equal to one. π₯ cubed minus π¦ cubed is then two
cubed minus one cubed, which is equal to seven. Then, π₯ plus π¦ times π₯ minus π¦
times π₯ plus π¦ is two plus one times two minus one times two plus one, which is
equal to nine.

Remember, since identities are true
for all values of the variables, then by showing that there is one set of variable
where this equation isnβt true, we see that we donβt have an identity. The answer then is no, itβs not an
identity.