Question Video: Evaluating Expressions Involving Cubic Roots of Unity | Nagwa Question Video: Evaluating Expressions Involving Cubic Roots of Unity | Nagwa

# Question Video: Evaluating Expressions Involving Cubic Roots of Unity Mathematics • Third Year of Secondary School

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What are the possible values of (8 + π)/(8πΒ² + 1) β πΒ², where π is a primitive cube root of unity?

03:05

### Video Transcript

What are the possible values of eight plus π over eight π squared plus one minus π squared, where π is a primitive cube root of unity?

First of all, let us remember that a cube root of unity is a number, π, such that π cubed equals one. Additionally, the root is specified to be primitive, which means in this instance that π cannot be one. In fact, we know explicitly that the two possible values that π can take are negative a half plus the square root of three over two π or negative a half minus the square root of three over two π. Let us denote these values by π one and π two.

One possible way to solve this problem would be to take these values and directly substitute them into the given expression. We could do this by making use of the fact that π one squared is π two and π two squared is π one. However, even though we would eventually get to the answer, this approach would not be ideal since we would have to do a lot of simplification. Therefore, we should first try and see whether there is a simpler method.

Let us start off by examining the given expression and seeing whether it can be simplified in any way. One thing we might notice is that the numerator of the fraction has the coefficients eight and one, while the denominator also has the coefficients eight and one. This suggests that the two halves of the fraction may be related somehow. This is in fact the case, and to demonstrate this, we can use the fact that by definition, π cubed equals one. So we can add an π cubed term to the first term of the numerator since itβs the same as multiplying by one.

Next, we can factor out π from the terms in the numerator since it is a common factor. And then we notice that eight π squared plus one can actually be canceled from the numerator and the denominator. So we are left with just π minus π squared.

Now we have two cases for the value of this expression depending on whether we substitute π one or π two. For π one, which is negative a half plus root three over two π, we have π one minus π one squared, which, due to the fact that π one squared is π two, is equal to π one minus π two, which is negative a half plus root three over two π minus negative a half minus root three over two π. And we can see that the real parts will cancel and the imaginary parts will combine together, giving us just the square root of three times π.

In the second case, we substitute π two for π. Thus, we have π two minus π two squared, which is π two minus π one. And this is just the negative of our first answer. Therefore, we have two possible values for the given expression, which are root three π and negative root three π.

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