Video: Finding the Coefficient of a Certain Degree of the Variable in a Binomial Expansion Product

Find the coefficient of π‘Žβ΅ in the expansion of (π‘ŽΒ² + (1/π‘ŽΒ²))Β³(π‘Ž + (1/π‘Ž))Β³.

06:31

Video Transcript

Find the coefficient of π‘Ž to the power of five in the expansion of π‘Ž squared plus one over π‘Ž squared to the power of three multiplied by π‘Ž plus one over π‘Ž to the power of three.

So in order to actually find out what the coefficient of π‘Ž to the power of five is in our expansion, what I’m first gonna do is expand each of the parentheses separately. Well, I’m gonna start with π‘Ž squared plus one over π‘Ž squared cubed. And to expand this, we’re actually gonna use the binomial theorem. And what the binomial theorem tells us is that if we have π‘Ž plus 𝑏 to the power of 𝑛, then this is equal to 𝑛 choose zero π‘Ž to the power of 𝑛 𝑏 to the power of zero plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 to the power of one, et cetera, carrying on this pattern up to 𝑛 choose 𝑛 π‘Ž to the power of zero 𝑏 to the power of 𝑛. Okay, great. So let’s apply this and actually find out what our expansion is going to be.

So our expansion gives us three choose zero π‘Ž squared to the power of three plus three choose one π‘Ž squared to the power of two multiplied by one over π‘Ž squared plus three choose two π‘Ž squared multiplied by one over π‘Ž squared to the power of two plus three choose three one over π‘Ž squared to the power of three. Okay, so now what we want to do is actually work out the coefficients of each of our terms. And to do this, we actually have, obviously, three choose one or three choose two, et cetera. And to find this out, we actually just put it into our calculator.

For example, if I want to define three choose one, I press three. And then press the nCr button on the calculator and then one. And that would give us our answer which would just be three. And we can also check our coefficients by using Pascal’s triangle. As we see that in the row that we’re looking at, and which will be the fourth row down because that’s when we’d have to the power of three, we’d have our coefficients as one, three, three, one. And so when we’ve done this on our calculator, we can check then that it actually meets this. Okay, so now let’s simplify.

So we get π‘Ž to the power of six plus three is our coefficient π‘Ž to the power of four over π‘Ž to the power of two plus three π‘Ž squared over π‘Ž to the power of four plus one over π‘Ž to the power of six. So now, we’re actually gonna simplify this. So we get π‘Ž to the power of six plus three π‘Ž squared. And that’s because if we have three π‘Ž to the power of four divided by π‘Ž squared, we actually subtract the powers. So four minus two gives us two plus three π‘Ž to the power of minus two plus π‘Ž to the power of minus six. And just to remind us how we actually got that last term, again, it’s one of our exponent rules. Well, this one is the one over π‘Ž to the power of 𝑏 equals π‘Ž to the power of negative 𝑏.

Okay, great. So we’ve now done that and we’ve expanded our first parentheses. Now, we’re gonna move on to the second one. So we’re gonna get π‘Ž to the power of three plus three π‘Ž squared multiplied by one over π‘Ž. And just to remind ourselves how we got that, well actually, our coefficient, we already know because we saw from the previous one that actually we work out our binomial coefficients. It gave us one, three, three, one. But also, from Pascal’s triangle, we know we get one, three, three, one. So therefore, the coefficient of this term is gonna be three. And then, we know that the exponent of π‘Ž is going to be two because it’s 𝑛 minus one because it’s our second term. And our exponent of one over π‘Ž is just gonna be one. Okay, great, so just reminded us how we got that. Let’s move on to the next terms.

So the next term’s gonna be plus three π‘Ž multiplied by one over π‘Ž squared. And then finally, plus one over π‘Ž cubed. Okay, great. So now, what we’re gonna do is simplify. So this is gonna lead us to π‘Ž cubed plus three π‘Ž squared over π‘Ž plus three π‘Ž over π‘Ž squared plus π‘Ž to the power of negative three. And then, as we did in the first step, we’re actually using an exponent rule here to actually simplify further because we’ve got π‘Ž to the power of 𝑏 over π‘Ž to the power of 𝑐 equals π‘Ž to the power of 𝑏 minus 𝑐. And this gives us π‘Ž cubed plus three π‘Ž. And we got that because we had three π‘Ž squared over π‘Ž, which is like π‘Ž to the power of one, so two minus one leaves us with one, so just three π‘Ž. And then, plus three π‘Ž to the power of negative one. And then, plus π‘Ž to the power of negative three.

Okay, so now we’ve expanded both. What do we do? Well, if we look back at the question, we were interested in the coefficient of π‘Ž to the power of five. So therefore, if we take a look at another exponent rule, which is π‘Ž to the power of 𝑏 multiplied by π‘Ž to the power of 𝑐 equals π‘Ž to the power of 𝑏 plus 𝑐, then we’re actually gonna use this to work out which of our terms are actually gonna give us π‘Ž to the power of five. And we can do this because if we look at the original expression, we can see that it actually gives our first set of parentheses multiplied by the second set of parentheses. So therefore, each of the terms is gonna be multiplied by each other.

So therefore, the first pair of terms we’re gonna look at is three π‘Ž to the power of two multiplied by π‘Ž to the power of three because this is gonna give us three π‘Ž to the power of five because if you’ve got the exponent two added to the exponent three, we get five. So, great. So that’s our first term that involves π‘Ž to the power of five. Then our next pair of terms we’re gonna look at is π‘Ž to the power of six multiplied by three π‘Ž to the power of negative one. And this is because if we have six add negative one, this is gonna give us five. So therefore, π‘Ž to the power of six multiplied by three π‘Ž to the power of negative one gives us three π‘Ž to the power of five.

And then, if we take a look at each of the combinations, this is the only two pairs of terms that actually multiply together to give us π‘Ž to the power of five. So therefore, we have three π‘Ž to the power of five and three π‘Ž to the power of five. So therefore, we’re gonna add these terms together to give us our term with π‘Ž to the power of five. And this gives us three π‘Ž to the power of five plus three π‘Ž to the power of five which equals six π‘Ž to the power of five.

And then, we check back at the question. And the question says, well, what’s the coefficient of π‘Ž to the power of five? So therefore, we can say that the coefficient of π‘Ž to the power of five is equal to six.

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