# Video: Proving Polynomial Identities

Is the equation π₯Β³ β π¦Β³ = (π₯ + π¦)(π₯ β π¦)(π₯ + π¦) an identity?

02:59

### 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.
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