Question Video: Proving Polynomial Identities Mathematics

Is the equation π‘₯Β³ βˆ’ 𝑦³ = (π‘₯ + 𝑦)(π‘₯ βˆ’ 𝑦)(π‘₯ + 𝑦) an identity?

03:31

Video Transcript

Is the equation π‘₯ cubed minus 𝑦 cubed equals π‘₯ plus 𝑦 multiplied by π‘₯ minus 𝑦 multiplied by π‘₯ plus 𝑦 an identity?

Well, for it to be an identity, the left-hand side must be the same as the right-hand side. So, what I’m gonna do is distribute across the parentheses in our right-hand side to see if it is, in fact, the same as the left-hand side of our equation. When we’re distributing across three sets of parentheses, what we do is we deal with two first of all. And then we distribute across the result and the last parentheses at the end. So, we’re gonna start with π‘₯ plus 𝑦 multiplied by π‘₯ minus 𝑦. So, what we want to do is multiply each term by each other. So, first of all, we’re gonna have π‘₯ multiplied by π‘₯, which gives us π‘₯ squared. Then, we’re gonna have π‘₯ multiplied by negative 𝑦, which gives us negative π‘₯𝑦. And then, we have 𝑦 multiplied by π‘₯, which gives us plus π‘₯𝑦. And then finally, positive 𝑦 multiplied by negative 𝑦, which gives us negative 𝑦 squared.

Well, we just multiplied each of the terms in the left-hand parentheses by each of the terms in the right-hand parentheses. But we can use a little memory aid, and that is FOIL, which means first, so we multiply the first terms together, outer, multiply the outer terms, then inner, and then last. Okay great, can we simplify now? Well, we have minus π‘₯𝑦 plus π‘₯𝑦. So, this is gonna be equal to zero. So, the result is going to be π‘₯ squared minus 𝑦 squared. And we could’ve written this straightaway from the off because we can see that it’s a formation of the difference of two squares because we have π‘₯ and then we have plus and then 𝑦 and then π‘₯ minus 𝑦. Because the signs are both different and because the final term is the same in both of our parentheses, we could’ve used the method just to write straightaway π‘₯ squared minus 𝑦 squared.

Okay, great. So now, let’s complete the last part of our distribution across our parentheses. So, what we’re gonna have is π‘₯ squared minus 𝑦 squared. Then, this is multiplied by π‘₯ plus 𝑦. So then, we’re gonna have π‘₯ squared multiplied by π‘₯, which gives us π‘₯ cubed. And then, we have π‘₯ squared multiplied by positive 𝑦, which gives us positive π‘₯ squared 𝑦. And then, we have negative 𝑦 squared multiplied by π‘₯, which gives us negative π‘₯𝑦 squared. And then finally, negative 𝑦 squared multiplied by positive 𝑦 will give us negative 𝑦 cubed. So, this is fully distributed now.

But this time, the two middle terms can’t cancel because we’ve got π‘₯ squared 𝑦 minus π‘₯𝑦 squared. So, we can see that the squareds are not the same because, in the first term, it’s π‘₯ squared, and in the second term, it’s the 𝑦 that’s squared. So therefore, we can see that π‘₯ cubed minus 𝑦 cubed is not identical to π‘₯ cubed plus π‘₯ squared 𝑦 minus π‘₯𝑦 squared minus π‘₯ cubed. So therefore, we’ve solved the problem. And we can say that, in answer to the question, β€œIs the equation π‘₯ cubed minus 𝑦 cubed equal to π‘₯ plus 𝑦 multiplied by π‘₯ minus 𝑦 multiplied by π‘₯ plus 𝑦 an identity?,” the answer is no.

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