Cube Roots Again!

Algebra Level 3

Find the value of ( 2 − ω ) ( 2 − ω 2 ) ( 2 − ω 10 ) ( 2 − ω 11 ) . (2-\omega)(2-\omega^2)(2-\omega^{10})(2-\omega^{11}).

Details and Assumptions:

ω \omega is a non-real cube root of unity.


The answer is 49.

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Parth Lohomi by Parth Lohomi , 20, India

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3 solutions

Discussions for this problem are now closed

As ω \omega is the cube root of unity.

⇒ ω 10 = ω , ω 11 = ω 2 \Rightarrow \omega^{10} = \omega, \omega^{11} = \omega^2

Then, ( 2 − ω ) ( 2 − ω 2 ) ( 2 − ω 10 ) ( 2 − ω 11 ) (2- \omega)(2- \omega^2)(2- \omega^{10})(2- \omega^{11})

= ( ( 2 − ω ) ( 2 − ω 2 ) ) 2 = ((2- \omega)(2- \omega^2))^2

Now, since ( x − ω ) ( x − ω 2 ) = x 2 + x + 1 (x- \omega)(x- \omega^2) = x^2 + x + 1 ,

= ( 2 2 + 2 + 1 ) 2 = (2^2 + 2 +1)^2

= 7 2 = 49 = 7^2 = \boxed{49} .

36 Helpful 0 Interesting 1 Brilliant 0 Confused

I liked how you use ( x − ω ) ( x − ω 2 ) = x 2 + x + 1 ( x - \omega) ( x - \omega^2 ) = x^2 + x + 1 . In cases like this, evaluating f ( x ) f(x) and then substituting in x = 2 x = 2 , is an easier approach than trying to figure out f ( 2 ) f(2) directly.

Calvin Lin Staff - 6 years, 6 months ago
Sandeep Bhardwaj
Dec 8, 2014

ω \omega being cube-root of unity. ( ≠ 1 ) \left( \neq 1 \right) , ω = − 1 + 3 i 2 \omega=\dfrac{-1+\sqrt{3}i}{2} or ω = − 1 − 3 i 2 \omega=\dfrac{-1-\sqrt{3}i}{2}

Here we will use the cute property of ω \omega : 1 + ω + ω 2 = 0 \boxed{1+\omega+\omega^2=0} So, ( 2 − ω ) ( 2 − ω 2 ) ( 2 − ω 10 ) ( 2 − ω 11 ) (2-\omega)(2-\omega^2)(2-\omega^{10})(2-\omega^{11})

= ( 2 − ω ) ( 2 − ω 2 ) ( 2 − ω ) ( 2 − ω 2 ) \quad = (2-\omega)(2-\omega^2)(2-\omega)(2-\omega^2) ( b e c a u s e ω 10 = ω , ω 11 = ω 2 ) \left( because \ \omega^{10}=\omega, \omega^{11}=\omega^2 \right)

= ( ( 2 − ω ) ( 2 − ω 2 ) ) 2 \quad = \left( (2-\omega)(2-\omega^2) \right)^2

= ( 4 − 2. ω − 2. ω 2 + ω 3 ) 2 \quad = \left( 4-2.\omega-2.\omega^2+\omega^3 \right)^2

= ( 5 − 2. ω − 2 ω 2 ) 2 \quad =\left( 5-2.\omega-2\omega^2 \right)^2 ( b e c a u s e ω 3 = 1 ) \left( because \ \omega^3=1 \right)

= ( 7 − 2 − 2. ω − 2. ω 2 ) 2 \quad =\left( 7-2-2.\omega-2.\omega^2 \right)^2

= ( 7 − 0 ) 2 = 49 \quad = \left( 7-0 \right)^2=\boxed{49}

26 Helpful 1 Interesting 1 Brilliant 0 Confused

Awesome solution, I will add more to your solution, demonstrating the cute property : 1 + ω + ω 2 = 0 \boxed{1+\omega+\omega^2=0}

Let ω 3 = 1 \omega^3=1

ω 3 − 1 = 0 \omega^3-1=0

( ω − 1 ) ( ω 2 + ω + 1 ) = 0 (\omega-1)(\omega^2+\omega+1)=0

So, or ω = 1 \omega=1 or ω 2 + ω + 1 = 0 \omega^2+\omega+1=0

Since ω \omega can't be 1

1 + ω + ω 2 = 0 \boxed{1+\omega+\omega^2=0}

Pedro Souto - 6 years, 6 months ago
Ariel Gershon
Dec 10, 2014

First since w 3 = 1 w^3 = 1 , we can rewrite the expression as ( 2 − w ) 2 ( 2 − w 2 ) 2 (2-w)^2(2-w^2)^2 .

Since w , w 2 , 1 w, w^2, 1 are the three roots of the equation x 3 − 1 = 0 x^3-1=0 , then that means ( x − 1 ) ( x − w ) ( x − w 2 ) = x 3 − 1 (x-1)(x-w)(x-w^2) = x^3 - 1 . If we let x = 2 x = 2 , we get ( 2 − 1 ) ( 2 − w ) ( 2 − w 2 ) = 2 3 − 1 (2-1)(2-w)(2-w^2) = 2^3 - 1 ( 2 − w ) ( 2 − w 2 ) = 7 (2-w)(2-w^2) = 7 ( 2 − w ) 2 ( 2 − w 2 ) 2 = 49 (2-w)^2(2-w^2)^2 = \boxed{49}

13 Helpful 2 Interesting 2 Brilliant 1 Confused

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