Clever Manipu 5 ^5 lation

Algebra Level 3

u 5 + 1 u 5 \Large u^5+\dfrac{1}{u^5} If u 2 + 1 u 2 = 14 u^2+\dfrac{1}{u^2}=14\; and u > 0 \;u>0 , then find the value of the above expression.


The answer is 724.

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Bloons Qoth by Bloons Qoth , 21, USA

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2 solutions

Note first that ( u + 1 u ) 2 = u 2 + 1 u 2 + 2 = 14 + 2 = 16 ⟹ u + 1 u = 4 \left(u + \dfrac{1}{u}\right)^{2} = u^{2} + \dfrac{1}{u^{2}} + 2 = 14 + 2 = 16 \Longrightarrow u + \dfrac{1}{u} = 4

since we are given that u > 0 u \gt 0 . Then

u 3 + 1 u 3 = ( u + 1 u ) ( u 2 + 1 u 2 ) − ( u + 1 u ) = 4 × 14 − 4 = 52 u^{3} + \dfrac{1}{u^{3}} = \left(u + \dfrac{1}{u}\right)\left(u^{2} + \dfrac{1}{u^{2}}\right) - \left(u + \dfrac{1}{u}\right) = 4 \times 14 - 4 = 52 .

So finally u 5 + 1 u 5 = ( u 2 + 1 u 2 ) ( u 3 + 1 u 3 ) − ( u + 1 u ) = 14 × 52 − 4 = 724 u^{5} + \dfrac{1}{u^{5}} = \left(u^{2} + \dfrac{1}{u^{2}}\right)\left(u^{3} + \dfrac{1}{u^{3}}\right) - \left(u + \dfrac{1}{u}\right) = 14 \times 52 - 4 = \boxed{724} .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

+1 Your solution is a great read:

  1. Nice choice of equations used to arrive at u 5 + 1 u 5 u^5 + \frac{1}{u^5 } .

Thanks for contributing and helping other members aspire to be like you

Calvin Lin Staff - 4 years, 8 months ago

I did the same way..

Fidel Simanjuntak - 4 years, 5 months ago
Tapas Mazumdar
Oct 14, 2016

u 2 + 1 u 2 = 14 \color{#3D99F6}{u^2 + \dfrac{1}{u^2} = 14}

Now, we have

u 5 + 1 u 5 = u 5 + 1 u 5 + ( u + 1 u ) − ( u + 1 u ) = u 5 + 1 u + u + 1 u 5 − ( u + 1 u ) = u 2 ( u 3 + 1 u 3 ) + 1 u 2 ( u 3 + 1 u 3 ) − ( u + 1 u ) = ( u 2 + 1 u 2 ) ( u 3 + 1 u 3 ) − ( u + 1 u ) = ( u 2 + 1 u 2 ) { ( u + 1 u ) ( u 2 + 1 u 2 − 1 ) } − ( u + 1 u ) = ( u 2 + 1 u 2 ) { ( ( u + 1 u ) 2 ) ( u 2 + 1 u 2 − 1 ) } − ( ( u + 1 u ) 2 ) = ( u 2 + 1 u 2 ) { ( u 2 + 1 u 2 + 2 ) ( u 2 + 1 u 2 − 1 ) } − ( u 2 + 1 u 2 + 2 ) = ( 14 ) { ( 14 + 2 ) ( 14 − 1 ) } − ( 14 + 2 ) = 14 ( 4 × 13 ) − 4 = 728 − 4 = 724 \begin{aligned} u^5 + \dfrac{1}{u^5} &= u^5 + \dfrac{1}{u^5} + \left( u + \dfrac 1u \right) - \left( u + \dfrac 1u \right) \\ &= u^5 + \dfrac 1u + u + \dfrac{1}{u^5} - \left( u + \dfrac 1u \right) \\ &= u^2 \left( u^3 + \dfrac{1}{u^3} \right) + \dfrac{1}{u^2} \left( u^3 + \dfrac{1}{u^3} \right) - \left( u + \dfrac 1u \right) \\ &= \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}}\right) \left( u^3 + \dfrac{1}{u^3} \right) - \left( u + \dfrac 1u \right) \\ &= \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}}\right) \left\{ \left( u + \dfrac 1u \right) \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}} - 1 \right) \right\} - \left( u + \dfrac 1u \right) \\ &= \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}} \right) \left\{ \left( \sqrt{ {\left( u + \dfrac 1u \right)}^2 } \right) \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}} - 1 \right) \right\} - \left( \sqrt{ {\left( u + \dfrac 1u \right)}^2 } \right) \\ &= \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}} \right) \left\{ \left( \sqrt{ {\color{#3D99F6}{ u^2 + \dfrac{1}{u^2}}} +2 } \right) \left( {\color{#3D99F6}{u^2 + \dfrac{1}{u^2}}} - 1 \right) \right\} - \left( \sqrt{ {\color{#3D99F6}{ u^2 + \dfrac{1}{u^2}}} +2 } \right) \\ &= \left( {\color{#3D99F6}{14}} \right) \left\{ \left( \sqrt{ {\color{#3D99F6}{14}} + 2} \right) \left( {\color{#3D99F6}{14}} - 1 \right) \right\} - \left( \sqrt{{\color{#3D99F6}{14}} + 2} \right) \\ &= 14 \left( 4 \times 13 \right) - 4 \\ &= 728 - 4 \\ &= \boxed{724} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

+1 This is correct! Given the slightly long writeup of your steps, it is worth considering whether a shorter solution exists.

Hint: ( x 2 + 1 x 2 ) ( x 3 + 1 x 3 ) = ( x 5 + 1 x 5 ) + ( x + 1 x ) \left (x^2 + \frac1{x^2}\right) \left(x^3 + \dfrac1{x^3} \right) = \left(x^5 + \dfrac1{x^5}\right) + \left( x + \dfrac1x \right) .

Pi Han Goh - 4 years, 8 months ago

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Actually I showed the factorization as well that's why there were three more steps to the solution. I think if you already know the equation that you've mentioned, it's fairly short to solve. I stuck to the long form to show how it's factorised.

Tapas Mazumdar - 4 years, 8 months ago

WOW!!! You are really good at factoring.I could never do that.

Razzi Masroor - 4 years, 7 months ago

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