### Video Transcript

Use the binomial theorem to expand two π₯ minus three π¦ cubed.

Weβre told how to answer this question. Weβre told to use the binomial theorem. Now, the binomial theorem is used to raise binomials to positive integer powers of π. It says that π plus π to the πth power for positive integer values of π is the sum from π equals zero to π of π choose π times π to the power of π minus π times π to the power of π. Now, that can be a little tricky to work with. So, we sometimes use the expanded form which says itβs equal to π to the πth power plus π choose one π to the power of π minus one π plus π choose two π to the power of π minus two π squared and so on all the way up to π to the πth power.

Now, weβre looking to expand two π₯ minus three π¦ cubed. So, weβll let π be equal to two π₯. π must be equal to negative three π¦. And of course, π, our exponent, is three. The first term is π to the πth power, so thatβs two π₯ cubed. We then have π choose one. Well, π is three here, so thatβs three choose one, times two π₯ squared times π, which is negative three π¦. Our third term is three choose two times two π₯ times negative three π¦ all squared. And then, our final term is π to the πth power. So, thatβs negative three π¦ cubed. Notice here that the powers of two π₯ are reducing by one each time and the powers of negative three π¦ are increasing by one each time.

Letβs simplify a little. Each time we have a term in brackets raised to some power, weβre going to distribute that power or exponent over each part of the term. So, two π₯ cubed is two cubed π₯ cubed or eight π₯ cubed. But what about our next term? We have three choose one there. And so we recall that π choose π is π factorial over π factorial times π minus π factorial. This means three choose one is three factorial over one factorial times three minus one factorial, or simply three factorial over one factorial times two factorial. But we recall that three factorial is actually three times two times one and two factorial is two times one. And then we notice we can divide the numerator and denominator of this fraction by two times one or two factorial.

And we therefore see that three choose one is three divided by one, which is three. So, our second term is three times four π₯ squared times negative three π¦. And that simplifies to negative 36π₯ squared π¦. Three choose two is three factorial over two factorial times one factorial. But we, of course, just worked out three factorial over one factorial times two factorial to be equal to three. So, three choose two must also be three. So, our third term is then three times two π₯ times negative three π¦ squared. And a negative times a negative is a positive, which is nine π¦ squared. And so this simplifies to 54π₯π¦ squared. Then, cubing a negative number yields a negative result. So, our fourth term is negative 27π¦ cubed.

And so, we found the binomial expansion of two π₯ minus three π¦ cubed. Itβs eight π₯ cubed minus 36π₯ squared π¦ plus 54π₯π¦ squared minus 27π¦ cubed.