A sleep into Mathematical Madness

Algebra Level 4

If a complex number α \alpha satisfies the equation α 3 − α 2 − 2 α + 1 = 0 , \alpha^3-\alpha^2-2\alpha+1=0, where α = x + 1 x \alpha=x+\frac{1}{x} for some complex number x x , then what is the value of the expression below? x 64 − 2 x 52 + 3 x 43 + 2 x 38 − 2 x 29 + 5 x 17 + 5 x 10 − 7 x 7 + 7 x^{64}-2x^{52}+3x^{43}+2x^{38}-2x^{29}+5x^{17}+5x^{10}-7x^7+7


The answer is 14.

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Sanjeet Raria by Sanjeet Raria , 27, India

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2 solutions

Sanjeet Raria
Dec 26, 2014

We have α 3 − α 2 − 2 α + 1 = 0 \large \alpha^3-\alpha^2-2\alpha+1=0

Now plugging α = ( x + 1 x ) \alpha=(x+\frac{1}{x}) we finally get ( x 3 + 1 x 3 ) − ( x 2 + 1 x 2 ) + ( x + 1 x ) − 1 = 0 (x^3+\frac{1}{x^3})- (x^2+\frac{1}{x^2})+ (x+\frac{1}{x})-1=0\space Multiplying throughout by x 3 x^3 we get x 6 − x 5 + x 4 − x 3 + x 2 − x + 1 = 0 x^6-x^5+x^4-x^3+x^2-x+1=0 Multiplying again throughout by ( x + 1 ) (x+1) we get, ( x + 1 ) ( x 6 − x 5 + x 4 − x 3 + x 2 − x + 1 ) = 0 (x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0 ⇒ x 7 + 1 = 0 ⇒ x 7 = − 1 \Rightarrow x^7+1=0 \Rightarrow x^7=-1

Now using this in our required expression x 64 − 2 x 52 + 3 x 43 + 2 x 38 − 2 x 29 + 5 x 17 + 5 x 10 − 7 x 7 + 7 x^{64}-2x^{52}+3x^{43}+2x^{38}-2x^{29}+5x^{17}+5x^{10}-7x^7+7 we simply it as − x + 2 x 3 + 3 x − 2 x 3 − 2 x + 5 x 3 − 5 x 3 − 7 ( − 1 ) + 7 = 14 -x+2x^3+3x-2x^3-2x+5x^3-5x^3-7(-1)+7=\boxed{14}

80 Helpful 16 Interesting 14 Brilliant 6 Confused

Nice solution. I did the same way. Keep it up Pumba. :)

Sandeep Bhardwaj - 6 years, 5 months ago

This question is beautiful.

Julian Poon - 6 years, 5 months ago

Nicely explained.I did the same.

Gautam Sharma - 6 years, 4 months ago

I am so stupid! I did the same thing but I misread the question. I took the coefficient of alpha^2 in the first equation as +1 instead of -1 and complicated up the whole thing >_<

Aditya Pappula - 5 years, 11 months ago

I did the same way as you did in both madnesses @Sanjeet Raria

Aakash Khandelwal - 5 years, 10 months ago

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Well done! I'm gonna post some more madness problems. You might like solving them too.

Sanjeet Raria - 5 years, 10 months ago

didn't get the multiplying by (x+1) part, wasn't able to think of it. smart thinking!

Pil Pinas - 4 years, 12 months ago

Isn't it supposed.to be (x + 1/x)^3 and not x^3 + 1/x^3

Ramesh Iyer - 3 months, 3 weeks ago
Julian Poon
Dec 30, 2014

This is a detailed elaboration on Sanjeet Raria's solution on why ( x 3 + 1 x 3 ) − ( x 2 + 1 x 2 ) + ( x + 1 x ) − 1 = 0 \left( { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } \right) -\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) +\left( { x }+\frac { 1 }{ { x } } \right) -1=0 First, lets expand ( x + 1 x ) 2 { \left( x+\frac { 1 }{ x } \right) }^{ 2 } and ( x + 1 x ) 3 { \left( x+\frac { 1 }{ x } \right) }^{ 3 } .

( x + 1 x ) 3 = x 3 + 1 x 3 + 3 ( x + 1 x ) = α 3 ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 = α 2 { \left( x+\frac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3\left( { x }+\frac { 1 }{ { x } } \right) ={ \alpha }^{ 3 }\\ { \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ \alpha }^{ 2 }

I hope you know where this is going. Now, we are going to express the polynomial in the question: α 3 − α 2 − 2 α + 1 { \alpha }^{ 3 }-{ \alpha }^{ 2 }-2{ \alpha }+1 in terms of x 3 + 1 x 3 , x 2 + 1 x 2 , and x + 1 x { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } }, \quad { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } , \text{ and } { x }+\frac { 1 }{ { x } }

So, x 2 + 1 x 2 = α 2 − 2 x 3 + 1 x 3 = α 3 − 3 α { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ \alpha }^{ 2 }-2\\ { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ={ \alpha }^{ 3 }-3{ \alpha } And therefore, ( x 3 + 1 x 3 ) − ( x 2 + 1 x 2 ) + ( x + 1 x ) − 1 = ( α 3 − 3 α ) − ( α 2 − 2 ) + α − 1 = α 3 − α 2 − 2 α + 1 = 0 \left( { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } \right) -\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) +\left( { x }+\frac { 1 }{ { x } } \right) -1 = ({ \alpha }^{ 3 }-3{ \alpha })-({ \alpha }^{ 2 }-2)+\alpha -1={ \alpha }^{ 3 }-{ \alpha }^{ 2 }-2{ \alpha }+1=\boxed{0}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

another way to find the equation for x x is : we have α x = x 2 + 1 \alpha x=x^2+1 and x 3 α 3 − x ( x α ) 2 − 2 x 2 ( x α ) + x 3 = 0 x^3\alpha^3-x(x\alpha)^2-2x^2(x\alpha)+x^3=0 so ( x 2 + 1 ) 3 − x ( x 2 + 1 ) 2 − 2 x 2 ( x 2 + 1 ) + x 3 = 0 (x^2+1)^3-x(x^2+1)^2-2x^2(x^2+1)+x^3=0

Cuize Han - 6 years, 5 months ago

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