Video: Evaluating Powers of Expressions Involving Cubic Roots of unity

Evaluate (2 + 2πœ” βˆ’ πœ”Β²)⁢ where πœ” is a cubic root of unity.

03:39

Video Transcript

Evaluate two plus two πœ” minus πœ” squared to the sixth power where πœ” is a cubic root of identity [unity]

This symbol is not a 𝑀 here. It’s the Greek letter πœ”. And it represents the cube root of one. And while one of the solutions is 𝑀 equal to one, it also has two other nonreal solutions, and all three of them satisfy 𝑀 cubed equals one.

There’s one other important property we need to know when working with this root of unity. And that’s that one plus πœ” plus πœ” squared equals zero. The first thing we wanna do when evaluating this expression is see if we can simplify or factor what’s inside the parentheses.

We notice that two of these terms have a factor of two. When we pull out that factor of two, we’ll be left with two times one plus πœ” minus πœ” squared to the sixth power. And when we see this one plus πœ”, it should remind us of this property we already know. One plus πœ” is equal to negative πœ” squared. In place of this one plus πœ”, we can substitute negative πœ” squared. We’ll then we have two times negative πœ” squared minus πœ” squared to the sixth power.

If we multiply two times negative πœ” squared, we’ll get negative two πœ” squared. And then, we need to subtract πœ” squared. When working with variables that are the same base to the same power, we can combine their coefficients. And that means we can combine negative two and negative one to make negative three πœ” squared to the sixth power. At this point, we wanna distribute this sixth power to the negative three and to the πœ” squared, negative three to the sixth power and πœ” squared to the sixth power.

We’ll need to remember the power-to-a-power rule. π‘₯ to the π‘Ž to the 𝑏 power equals π‘₯ to the π‘Ž times 𝑏. This means that we’ll be dealing with πœ” to the two times sixth power, πœ” to the 12th. Negative three to the sixth power equals 729. Now we have 729 times πœ” to the 12th power. But we know that πœ” cubed equals one. And we could write πœ” to the 12th power as πœ” cubed times itself four times. This is what happens when you multiply powers with the same base. You end up adding the exponents.

π‘₯ to the π‘Ž times π‘₯ to the 𝑏 equals π‘₯ to the π‘Ž plus 𝑏. πœ” to the 12th power is equal to πœ” cubed times πœ” cubed times πœ” cubed times πœ” cubed, which we could rewrite as πœ” cubed to the fourth power. Now we’ll have 729 times πœ” cubed to the fourth power. But we know that πœ” cubed equals one, and that one to the fourth power still equals one. 729 times one equals 729. So, the final solution of this expression is 729.

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