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Algebra Level 4

( 2 ω + 1 ω ) 1000 − ( ω 2 − 1 ω 2 ) 1000 = ? \large { { \left (\frac { 2\omega +1 }{ \omega } \right ) }^{ 1000 }-{ \left (\frac { { \omega }^{ 2 }-1 }{ { \omega }^{ 2 } } \right ) }^{ 1000 } = \ ? }

Details and Assumptions

w w denote the cube root unity, which is the root of x 3 = 1 x^3 = 1 other than 1 1 .


The answer is 0.000.

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Abdulrahman El Shafei by Abdulrahman El Shafei , 23, Egypt

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6 solutions

X = ( 2 ω + 1 ω ) 1000 − ( ω 2 − 1 ω 2 ) 1000 ∵ ω 3 = 1 ∴ X = ( ω 3 ( 2 ω + 1 ) ω ) 1000 − ( ω 3 ( ω 2 − 1 ) ω 2 ) 1000 X = ( ω 2 ( 2 ω + 1 ) ) 1000 − ( ω ( ω 2 − 1 ) ) 1000 X = ( 2 ω 3 + ω 2 ) 1000 − ( ω 3 − ω ) 1000 X = ( 2 + ω 2 ) 1000 − ( 1 − ω ) 1000 X = ( 1 + 1 + ω 2 ) 1000 − ( 1 − ω ) 1000 X = ( 1 − ω ) 1000 − ( 1 − ω ) 1000 = 0 \huge{ X={ (\frac { 2\omega +1 }{ \omega } ) }^{ 1000 }-{ (\frac { { \omega }^{ 2 }-1 }{ { \omega }^{ 2 } } ) }^{ 1000 }\\ { \because \omega }^{ 3 }=1\\ \therefore X={ (\frac { { \omega }^{ 3 }(2\omega +1) }{ \omega } ) }^{ 1000 }-{ (\frac { { { \omega }^{ 3 }(\omega }^{ 2 }-1) }{ { \omega }^{ 2 } } ) }^{ 1000 }\\ \quad X={ ({ \omega }^{ 2 }(2\omega +1)) }^{ 1000 }-{ ({ { \omega }(\omega }^{ 2 }-1)) }^{ 1000 }\\ \quad X\quad =\quad { (2{ \omega }^{ 3 }+{ \omega }^{ 2 }) }^{ 1000 }-{ ({ \omega }^{ 3 }-\omega ) }^{ 1000 }\\ \quad X\quad =\quad { (2+{ \omega }^{ 2 }) }^{ 1000 }-{ (1-\omega ) }^{ 1000 }\\ \quad X\quad =\quad { (1+1+{ \omega }^{ 2 }) }^{ 1000 }-{ (1-\omega ) }^{ 1000 }\\ \quad X\quad =\quad { (1-\omega ) }^{ 1000 }-{ (1-\omega ) }^{ 1000 }=\boxed { 0 } }

Conclusion: X = ( 2 ω + 1 ω ) n − ( ω 2 − 1 ω 2 ) n = 0 \huge{ \\ X={ (\frac { 2\omega +1 }{ \omega } ) }^{ n }-{ (\frac { { \omega }^{ 2 }-1 }{ { \omega }^{ 2 } } ) }^{ n }=0 }

22 Helpful 0 Interesting 0 Brilliant 0 Confused

How is (1+1+w^2) = 1-w? I don't get it. Please enlighten me.

Clifford Lesmoras - 6 years, 5 months ago

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We have as a result of roots of unity: 1 + ω + ω 2 = 0 1 + \omega + \omega^{2} = 0 . Thus 1 + ω 2 = − ω 1 + \omega^{2} = -\omega . From this we simplify ( 1 + ( 1 + ω 2 ) ) (1 + (1 + \omega^{2})) to ( 1 − ω ) (1 - \omega) .

Isay Katsman - 6 years, 5 months ago

Sorry,, I didn't explain it, actually I didn't see your comment except right now ,, @Isay Katsman has good explanation if you are still confused see the solutions of this problem

Abdulrahman El Shafei - 6 years, 5 months ago

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I actually computed it xD and gave me the exact same result, but your explanation is more elegant because I suspect some problems require your method.

Radinoiu Damian - 6 years, 5 months ago
Sujoy Roy
Dec 25, 2014

( 2 ω + 1 ω ) 1000 − ( ω 2 − 1 ω 2 ) 1000 (\frac{2\omega+1}{\omega})^{1000}-(\frac{\omega^2-1}{\omega^2})^{1000}

= ( 2 ω 3 + ω 2 ω 3 ) 1000 − ( ω 3 − ω ω 3 ) 1000 =(\frac{2\omega^3+\omega^2}{\omega^3})^{1000}-(\frac{\omega^3-\omega}{\omega^3})^{1000}

= ( 1 + 1 + ω 2 ) 1000 − ( 1 − ω ) 1000 =(1+1+\omega^2)^{1000}-(1-\omega)^{1000}

= ( 1 − ω ) 1000 − ( 1 − ω ) 1000 = 0 =(1-\omega)^{1000}-(1-\omega)^{1000}=\boxed{0}

9 Helpful 0 Interesting 0 Brilliant 0 Confused
Kartik Sharma
Dec 26, 2014

X = 2 ω + 1 ω 1000 − ω 2 − 1 ω 2 1000 X= {\frac{2\omega + 1}{\omega}}^{1000} - {\frac{{\omega}^{2} - 1}{{\omega}^{2}}}^{1000}

Let a = 2 ω + 1 ω , b = ω 2 − 1 ω 2 a = \frac{2\omega + 1}{\omega}, b = \frac{{\omega}^{2} -1}{{\omega}^{2}}

X = a 1000 − b 1000 X = {a}^{1000} - {b}^{1000}

X = ( a − b ) ( a 999 + a 998 b + . . . . . b 998 a + b 999 ) X = (a-b)({a}^{999} + {a}^{998}b +.....{b}^{998}a + {b}^{999})

a − b = 2 ω + 1 ω − ω 2 − 1 ω 2 a - b = \frac{2\omega + 1}{\omega} - \frac{{\omega}^{2} -1}{{\omega}^{2}}

= 2 ω 2 + ω − ω 2 + 1 ω 2 = \frac{2{\omega}^{2} + \omega - {\omega}^{2} +1}{{\omega}^{2}}

= ω 2 + ω + 1 ω 2 = \frac{{\omega}^{2} + \omega + 1}{{\omega}^{2}}

Now, using geometric sum formula,

a − b = ω 3 − 1 ω − 1 ω 2 a-b = \frac{\frac{{\omega}^{3} - 1}{\omega - 1}}{{\omega}^{2}}

We know ω 3 = 1 {\omega}^{3} = 1

Hence, a − b = 0 a-b = 0

Therefore, X = 0 X = 0

7 Helpful 0 Interesting 0 Brilliant 0 Confused

This helps. Thanks.

Clifford Lesmoras - 6 years, 5 months ago

how to use geometric sum frmula? give me link i wanna know math

Louiegie Paquit - 6 years, 5 months ago

Just 1+t+t²+t³+...+tⁿ= (1-t^(n+1))/(1-t)=(t^(n+1)-1)/(t-1) That is formula for geometric sum 1+x+x²+...+xⁿ

Nikola Djuric - 4 years, 4 months ago
Chew-Seong Cheong
Dec 27, 2014

The cubic roots of 1 satisfy the equation: ω 2 + ω + 1 = 0 \omega^2 + \omega + 1 =0 .

Therefore,

X = ( 2 ω + 1 ω ) 1000 − ( ω 2 − 1 ω 2 ) 1000 X = \left( \dfrac {2\omega+1}{\omega} \right)^{1000} - \left( \dfrac {\omega^2-1}{\omega^2} \right)^{1000}

= ( ω + ( ω + 1 ) ω ) 1000 − ( ( ω − 1 ) ( ω + 1 ) − ( ω + 1 ) ) 1000 \quad = \left( \dfrac {\omega + (\omega+1)}{\omega} \right)^{1000} - \left( \dfrac {(\omega-1)(\omega+1)}{-(\omega+1)} \right)^{1000}

= ( ω − ω 2 ω ) 1000 − ( 1 − ω ) 1000 \quad = \left( \dfrac {\omega - \omega^2}{\omega} \right)^{1000} - \left( 1-\omega \right)^{1000}

= ( 1 − ω ) 1000 − ( 1 − ω ) 1000 = 0 \quad = \left( 1 - \omega \right)^{1000} - \left( 1-\omega \right)^{1000} = \boxed{0}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
U Z
Dec 25, 2014

X = ( 2 + w 3 w ) 1000 − ( 1 − w 3 w 2 ) 1000 X = ( 2 + \dfrac{w^3}{w})^{1000} - ( 1 - \dfrac{w^3}{w^2})^{1000}

X = [ ( 2 + w 2 ) 2 ] 500 − [ ( 1 − w ) 2 ] 500 X = [(2 + w^2)^2]^{500} - [(1 - w)^2]^{500}

X = ( − 3 w ) 500 − ( − 3 w ) 500 = 0 X = ( -3w)^{500} - (-3w)^{500} = 0

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Aditya Kumar
Dec 26, 2014

Multiply the denominator and numerator of the first term with ω \omega . Note that:

1 + ω + ω 2 = 0 \huge{1+\omega+\omega^2=0} 2 ω + 1 ω = 2 ω 2 + ω ω 2 = ω 2 + ( − 1 − ω ) + ω ω 2 = ω 2 − 1 ω 2 \huge{\frac{2\omega+1}{\omega}=\frac{2\omega^2+\omega}{\omega^2}=\frac{\omega^2+(-1-\omega)+\omega}{\omega^2}=\frac{\omega^2-1}{\omega^2}}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

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