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Evaluate the square root of three plus one cubed plus the square root of three minus one cubed using the binomial expansion theorem.
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Weβll begin by recalling this theorem.
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For positive integer values of π, the binomial theorem tells us that π plus π to the πth power is given by 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 πth power.
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Now, that can be quite difficult to work with.
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So, we might consider, alternatively, its expanded form.
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Itβs 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 all the way up to π to the πth power.
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Notice how the exponent of π or the power of π reduces by one each time whereas the power for π increases by one each time.
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So, letβs begin with the square root of three plus one cubed.
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Weβre going to define π as root three; π is one.
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And weβre raising this to the third power, so π is equal to three.
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Our first term is π to the πth power.
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So, thatβs the square root of three cubed.
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Our second term is π choose one.
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Thatβs three choose one times the square root of three squared times one.
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We reduce the power of the square root of three by one and increase the power of one by one.
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So, our third term is three choose two root three times one squared.
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And then, our final term is one cubed.
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Weβre now going to evaluate each part of this expansion.
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Letβs write the square root of three cubed as the square root of three squared times the square root of three.
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Now, the square root of three squared is simply three.
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So, our first term is three root three.
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The square root of three squared times one is three, so our second term becomes three choose one times three.
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But what is three choose one?
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We recall that π choose π is π factorial over π factorial times π minus π factorial.
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This means three choose one is three factorial over one factorial times three minus one factorial or three factorial over one factorial times two factorial.
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We recall, though, that three factorial is three times two times one and two factorial is two times one.
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So, we see that we can divide through by two factorial, and three choose one is, therefore, three divided by one, which is three.
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So, our second term simplifies to three times three, which is nine.
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In fact, three choose two is also three.
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So, our third term is three root three.
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And our fourth term is one cubed, which is one.
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We see this simplifies really nicely to six root three plus 10.
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Now, weβre going to repeat this process for the square root of three minus one.
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Now notice this is almost identical to our earlier expansion, but this time π is equal to negative one.
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So, when we expand, we get root three cubed plus three choose one root three squared times negative one plus three choose two root three times negative one squared plus negative one cubed.
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This becomes three root three minus nine plus three root three minus one.
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And that simplifies to six root three minus 10.
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The question asks us to evaluate the sum of these expansions.
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So, thatβs six root three plus 10 plus six root three minus 10.
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Now, it should be quite clear that plus 10 minus 10 is zero.
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And so, we get six root three plus six root three, which is 12 root three.
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And so, weβre finished.
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The square root of three plus one cubed plus the square root of three minus one cubed is 12 root three.