### Video Transcript

Evaluate five plus five π plus seven π squared all raised to the power of negative four.

For this question, we can first recognize that the term π represents one of the complex cube roots of unity. The cube roots of unity are the solutions to the cube root of one. In other words, they are the numbers that are equal to one when they are cubed. The obvious solution to the cube root of one is one. However, you may be familiar with the other two solutions, which are minus one plus the square root of three times π all divided by two and minus one minus the square root of three π all divided by two.

For this video, we wonβt go into the calculation of these roots. However, you should be able to see the we have one real number and two complex numbers. In order to answer our question, we can first demonstrate an interesting property of these two complex roots. This property is that the square of either of the complex cubic roots of unity is equal to the other complex cube root of unity.

In order to prove this, letβs pick one of our complex cube roots of unity and define it as π. Letβs now take out a factor of a half and square each of the parentheses to find an equation for π squared. We can now simplify by saying that a half squared is a quarter in expanding out the terms of our binomial squared for the right-hand parentheses. Simplifying further, we find that we have a term of three π squared in our parentheses. To move forward, we can recall that the definition of π is the square root of negative one. π squared is therefore negative one and three π squared is equal to negative three.

Letβs sub in this value back to our equation. Doing this allows us to perform further simplifications. For our next step, we can take out a factor of two from our parentheses and finally we reach a result. Looking at this, we can see that the answer we found does indeed match the other complex cube root of unity.

Now, itβs worth noting that this relationship would still be true if weβd instead chosen our other complex cube root of unity as the original π. We wonβt go through the same calculation steps in this video. But you can do this on your own if you like. From this process, we can conclude that the three cubic roots of unity can be defined as one, π, and π squared.

Let us now return to the expression given in the question. The first thing we can do is to take out a factor of five from the first two terms in our parentheses. Let us now work on finding a meaningful substitution that will help us move forward. We have now defined π and π squared as our two complex cubic roots of unity. Since we know the value for both of these roots, weβre able to write out an equation for the sum π plus π squared. Splitting our two fractions into their individual terms allows us to simplify. And we find that π plus π squared is equal to negative one.

Now that weβve simplified this equation, letβs perform a rearrangement to make it more useful for our substitution. By first adding one to both sides of our equation and then subtracting π squared from both sides of our equation, we see that π plus one is equal to negative π squared. We can use this relationship to sub into our expression. Doing so, we find that our expression now only contains terms of π squared. Multiplying our factor of five back in and simplifying, we find that our parentheses reduce to two π squared.

Weβre now ready to raise the contents of our parentheses to the power of negative four, which gives us two to the power of negative four multiplied by π to the power of negative eight. Now, we can simplify two to the power of negative four as one over two to the power of four which is equal to one over 16. In order to simplify π to the power of negative eight, weβre gonna to have to work a little bit harder.

Now by definition, we know the cubing any of our cubic roots of unity will give us one. Since π is one of these roots, we know that π cubed is equal to one. Since this is true, letβs see what happens when we raise π cubed to any integer power π. Since one raised to the power of π is still equal to one, weβre able to say that π cubed raised to any power π will also still be equal to one. We can reduce this slightly and say that π to the power of three π is equal to one.

Given this fact, we know that multiplying any number by π to the power of three π is the same as multiplying by one and the number will remain unchanged. We can use this to simplify π to the power of negative eight by taking the case of π equals three. Here, we see that π to the power of nine is equal to one. Letβs now construct an equation.

We know that multiplying π to the power of negative eight by π to the power of nine is the same as multiplying π to the power of negative eight by one. To simplify, we can take the addition of the two powers on the left-hand side of our equation. And we should be able to see that π to the power of negative eight is equal to π. Letβs now use this as a substitution to simplify our original expression.

Performing the substitutions, we finally reach our answer. In doing so, we have completed our question. And we find that evaluating five plus five π plus seven π squared all raised to the power of negative four, we can simplify to one over 16 times π.