# Video: EG19M2-ALGANDGEO-Q17

Evaluate (π + ππ€ + ππ€Β²)(π + ππ€Β² + ππ€β΄).

03:29

### Video Transcript

Evaluate π plus ππ€ plus ππ€ squared times π plus ππ€ squared plus ππ€ to the power of four.

These π€s are related to the cube roots of unity. So, we will use some of these properties to rewrite our expression that way we can evaluate it. One of the properties is that π€ cubed is equal to one. So, with the cube root of unity, we know that π€ is a root of the equation π§ cubed minus one equals zero. Therefore, π€ satisfies the equation π§ cube minus one equals zero. Consequently, we can say π€ cubed minus one equals zero. Or adding one to both sides of the equation, we get π€ cubed equals one.

The sum of three cube root of unity is zero. One plus π€ plus π€ squared is equal to zero. So, we will use these two properties to help us evaluate. So, to evaluate, we donβt want to necessarily expand the brackets. We want to work inside the brackets and simplify. Letβs rearrange in the first set of brackets to put the πs together. So, we have π plus ππ€ squared, and then we have our plus ππ€. And we do the same for the second. This allows us to take out the π. And this will be helpful because inside the pink brackets, we can rewrite these expressions.

Notice, we have π€ squared plus one. In the top-right-hand corner, we have that one plus π€ plus π€ squared is equal to zero. So, if we would subtract π€ from both sides of the equation, we would have that one plus π€ squared would actually be equal to negative π€. And thatβs what we can do here. Now over here in this next set, the π€ to the power of four can be rewritten as π€ cubed times π€ to the power of one. But we know that π€ cubed is equal to one. And one times π€ to the power of one is simply π€. So, we can replace π€ to the power of four with π€.

However, this one plus π€ can be rewritten because we have one plus π€ in the top-right-hand corner. So, if we would subtract π€ squared from both sides of that equation, we would have that one plus π€ would be equal to negative π€ squared. So, letβs go ahead and replace both of these. So, now we have negative ππ€ plus ππ€ times negative ππ€ squared plus ππ€ squared.

We can first take out a π€. And then, in our second brackets, we can take out a π€ squared. But instead of writing negative π plus π, letβs go ahead and write it as π minus π simply because itβs more commonly written this way. Now, notice, we have π€π minus ππ€ squared and π minus π, and all four of those things are being multiplied together. So, if we took π€ times π€ squared, we will get π€ cubed. And π minus π times π minus π would be equal to π minus π squared. Now we also know what π€ cubed is equal to. Itβs equal to one. Therefore, we have one times π minus π squared. Therefore, after evaluating, we will find that this would be equal to π minus π squared.