(Easy) Radical Expression

Algebra Level 3

x = 4 ( 3 + 1 ) ( 3 4 + 1 ) ( 3 8 + 1 ) x=\frac{4}{\left(\sqrt{3}+1\right)\left(\sqrt[4]{3}+1\right)\left(\sqrt[8]{3}+1\right)}

Find the value of ( x + 2 ) 8 \left(x+2\right)^8


The answer is 768.

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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2 solutions

Chew-Seong Cheong
Jun 18, 2019

x = 4 ( 3 + 1 ) ( 3 4 + 1 ) ( 3 8 + 1 ) × 3 8 − 1 3 8 − 1 = 4 ( 3 8 − 1 ) ( 3 + 1 ) ( 3 4 + 1 ) ( 3 4 − 1 ) = 4 ( 3 8 − 1 ) ( 3 + 1 ) ( 3 − 1 ) = 4 ( 3 8 − 1 ) 3 − 1 = 2 ( 3 8 − 1 ) \begin{aligned} x & = \frac 4{(\sqrt 3+1)(\sqrt[4]3+1)(\sqrt[8]3+1)} \color{#3D99F6} \times \frac {\sqrt[8]3-1}{\sqrt[8]3-1} \\ & = \frac {4(\sqrt[8]3-1)}{(\sqrt 3+1)(\sqrt[4]3+1)(\sqrt[4]3-1)} \\ & = \frac {4(\sqrt[8]3-1)}{(\sqrt 3+1)(\sqrt 3-1)} \\ & = \frac {4(\sqrt[8]3-1)}{3-1} \\ & = 2(\sqrt[8]3-1) \end{aligned}

Therefore, ( x + 2 ) 8 = ( 2 ( 3 8 − 1 ) + 2 ) 8 = 2 8 ( 3 8 ) 8 = 2 8 × 3 = 768 (x+2)^8 = \left(2(\sqrt[8]3-1)+2\right)^8 = 2^8 \left(\sqrt[8]3 \right)^8 = 2^8 \times 3 = \boxed{768} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Chris Lewis
Jun 18, 2019

Repeatedly factoring using difference of two squares, we have

t 8 − 1 = ( t 4 − 1 ) ( t 4 + 1 ) = ( t 2 − 1 ) ( t 2 + 1 ) ( t 4 + 1 ) = ( t − 1 ) ( t + 1 ) ( t 2 + 1 ) ( t 4 + 1 ) t^8-1=\left( t^4-1\right) \left(t^4+1 \right)=\left( t^2-1\right)\left( t^2+1\right)\left( t^4+1\right) =\left(t-1 \right)\left( t+1\right)\left( t^2+1\right)\left( t^4+1\right)

With t = 3 8 t=\sqrt[8]{3} , the expression in the question is

x = 4 ( t + 1 ) ( t 2 + 1 ) ( t 4 + 1 ) = 4 ( t − 1 ) t 8 − 1 = 4 ( 3 8 − 1 ) 2 = 2 3 8 − 2 x=\frac{4}{\left(t+1\right)\left( t^2+1\right)\left( t^4+1\right) }\\ =\frac{4(t-1)}{t^8-1}\\ =\frac{4 \left(\sqrt[8]{3}-1 \right)}{2}\\ =2\sqrt[8]{3}-2

Hence ( x + 2 ) 8 = 2 8 ⋅ 3 = 768 (x+2)^8=2^8\cdot 3 = \boxed{768} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Any tips on formatting the LaTeX above would be appreciated! (In particular, alignment at equals signs and spacing between lines).

Chris Lewis - 1 year, 11 months ago

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Just add \begin{eqnarray} and \end{eqnarray} .

For example: x = 4 ( t + 1 ) ( t 2 + 1 ) ( t 4 + 1 ) = 4 ( t − 1 ) t 8 − 1 = 4 ( 3 8 − 1 ) 2 = 2 3 8 − 2 \begin{aligned} x&=&\frac{4}{\left(t+1\right)\left( t^2+1\right)\left( t^4+1\right) }\\ &=&\frac{4(t-1)}{t^8-1}\\ &=&\frac{4 \left(\sqrt[8]{3}-1 \right)}{2}\\ &=&2\sqrt[8]{3}-2 \end{aligned}

Or you might be interested in \begin{array} { l l l } and \end{array}

For example: x = 4 ( t + 1 ) ( t 2 + 1 ) ( t 4 + 1 ) = 4 ( t − 1 ) t 8 − 1 = 4 ( 3 8 − 1 ) 2 = 2 3 8 − 2 \begin{array} {lll} x&=&\frac{4}{\left(t+1\right)\left( t^2+1\right)\left( t^4+1\right) }\\ &=&\frac{4(t-1)}{t^8-1}\\ &=&\frac{4 \left(\sqrt[8]{3}-1 \right)}{2}\\ &=&2\sqrt[8]{3}-2 \end{array}

Note that the l l l represents left left left . Feel free to change any of the l s to r (right) or c (center).

Pi Han Goh - 1 year, 11 months ago

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Great, thank you!

Chris Lewis - 1 year, 11 months ago

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