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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Evaluate (9 β πΒ² + 9πβ΄)Β² + (6 + 6πΒ² + 6πβ΄)Β², where π is a nontrivial cubic root of unity.

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### Video Transcript

Evaluate nine minus π squared plus nine π to the fourth power all squared plus six plus six π squared plus six π to the fourth power all squared, where π is a nontrivial cubic root of unity.

In this question, we are asked to evaluate an expression involving π, which we are told is a nontrivial cubic root of unity.

To answer this question, we can start by noting that since π is a cube root of unity, the cube of π must be equal to one. We can multiply this equation through by π to see that π to the fourth power is equal to π. We can use this to rewrite the given expression. We want to replace every instance of π to the fourth power with π. Doing this gives us the following expression. We can simplify the expression further by recalling that the sum of the powers of a nontrivial root of unity is zero. So one plus π plus π squared equals zero.

We can take out the shared factor of six in the second term to obtain the following expression. We can then note that the factor of one plus π plus π squared is equal to zero. Therefore, this term is equal to zero, so we are left with only the first term. We can simplify even further by noting that there are two terms which share a factor of nine. We can then rewrite this factor in terms of π squared by noting that one plus π must be equal to negative π squared. We can substitute this into the expression to get nine times negative π squared minus π squared all squared. We can then simplify the expression inside the parentheses to obtain negative 10π squared all squared.

Now, we apply the laws of exponents to take the power of each factor separately. We have negative 10 squared times π squared squared, which is equal to π to the fourth power. Finally, we can use the fact that π to the fourth power is equal to π to simplify the expression to obtain a final answer of 100π.

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