The answer is an integer?

Algebra Level 3

2 + 4 − 2 2 3 + 2 − 4 − 2 2 3 \large \sqrt[3]{ 2 + \sqrt{4- 2\sqrt{2}}} + \sqrt[3]{ 2 - \sqrt{4- 2\sqrt{2}}}

If x x equals the expression above, then find the value of x 3 − 3 2 x + 4 x^{3} - 3\sqrt{2}x + 4 .


The answer is 8.

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Peter Bishop by Peter Bishop , 40, Singapore

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2 solutions

x = 2 + 4 − 2 2 3 + 2 − 4 − 2 2 3 x=\sqrt[3]{ 2 + \sqrt{4- 2\sqrt{2}}} + \sqrt[3]{ 2 - \sqrt{4- 2\sqrt{2}}}

Cube both sides:

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

x 3 = 4 + 3 ( 2 + 4 − 2 2 ) ( 2 − 4 − 2 2 ) 3 x x^3=4+3\sqrt[3]{(2 + \sqrt{4- 2\sqrt{2}})(2 - \sqrt{4- 2\sqrt{2}})}x

x 3 = 4 + 3 4 − ( 4 − 2 2 ) 3 x x^3=4+3\sqrt[3]{4-(4-2\sqrt{2})}x

x 3 = 4 + 3 8 3 x x^3=4+3\sqrt[3]{\sqrt{8}}x

x 3 = 4 + 3 2 x x^3=4+3\sqrt{2}x

x 3 − 3 2 x − 4 = 0 x^3-3\sqrt{2}x-4=0

Finally, add 8 8 to both sides:

x 3 − 3 2 x + 4 = 8 x^3-3\sqrt{2}x+4=\boxed{8}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

Bonus question : What are the other roots to the equation x 3 − 3 2 x − 4 = 0 x^3-3\sqrt{2}x-4=0 ?

If we consider the identity ( u + v ) 3 − 3 u v ( u + v ) − ( u 3 + v 3 ) = 0 (u+v)^3-3uv(u+v)-(u^3+v^3)=0 and make it match with the equation, we get the system:

x = u + v u v = 2 u 3 + v 3 = 4 x=u+v \\ uv=\sqrt{2} \\ u^3+v^3=4

By Vieta's formulas, the solutions for u u and v v are:

u , v = 2 ± 4 − 2 2 3 u,v=\sqrt[3]{2 \pm \sqrt{4-2\sqrt{2}}} .

But, ( u w , v w 2 ) (uw,vw^2) and ( u w 2 , v w ) (uw^2,vw) are also solutions of the system, where w w is a primitive cube root of unity.

So,

x 1 = 2 + 4 − 2 2 3 + 2 − 4 − 2 2 3 x 2 = 2 + 4 − 2 2 3 w + 2 − 4 − 2 2 3 w 2 x 3 = 2 + 4 − 2 2 3 w 2 + 2 − 4 − 2 2 3 w x_1=\sqrt[3]{2+\sqrt{4-2\sqrt{2}}}+\sqrt[3]{2 - \sqrt{4-2\sqrt{2}}} \\x_2=\sqrt[3]{2+\sqrt{4-2\sqrt{2}}}w+\sqrt[3]{2 - \sqrt{4-2\sqrt{2}}}w^2 \\x_3=\sqrt[3]{2+\sqrt{4-2\sqrt{2}}}w^2+\sqrt[3]{2 - \sqrt{4-2\sqrt{2}}}w

Alan Enrique Ontiveros Salazar - 5 years, 12 months ago

1.656235178 -0.828117589+0.5561405236i -0.828117589-0.5561405236i

Lee Isaac - 5 years, 12 months ago
Mohamed Wafik
Jun 16, 2015

assume x = ( 2 + 4 − 2 2 ) 1 / 3 + ( 2 − 4 − 2 2 ) 1 / 3 x = (2 + \sqrt{4 -2 \sqrt{2} } )^{1/3} + (2 - \sqrt{4 -2 \sqrt{2} } )^{1/3}

x 2 = ( 2 + 4 − 2 2 ) 2 / 3 + ( 2 − 4 − 2 2 ) 2 / 3 + 2 × ( 4 − ( 4 − 2 2 ) ) 1 / 3 x^{2} = (2 + \sqrt{4 -2 \sqrt{2} } )^{2/3} + (2 - \sqrt{4 -2 \sqrt{2} } )^{2/3} + 2 \times (4 - (4 - 2 \sqrt{2}))^{1/3}

taking the last term we find that

( 4 − ( 4 − 2 2 ) ) 1 / 3 = ( 2 2 ) 1 / 3 = 2 1 / 3 × 2 1 / 6 = 2 (4 - (4 - 2 \sqrt{2}))^{1/3} = (2 \sqrt{2})^{1/3} = 2 ^ {1/3} \times 2^{1/6} = \sqrt{2} then

x 2 = ( 2 + 4 − 2 2 ) 2 / 3 + ( 2 − 4 − 2 2 ) 2 / 3 + 2 2 x^{2} = (2 + \sqrt{4 -2 \sqrt{2} } )^{2/3} + (2 - \sqrt{4 -2 \sqrt{2} } )^{2/3} + 2 \sqrt{2}

x 2 − 3 2 = ( 2 + 4 − 2 2 ) 2 / 3 + ( 2 − 4 − 2 2 ) 2 / 3 − 2 x^{2} - 3 \sqrt{2} = (2 + \sqrt{4 -2 \sqrt{2} } )^{2/3} + (2 - \sqrt{4 -2 \sqrt{2} } )^{2/3} - \sqrt{2}

x × ( x 2 − 3 2 ) = ( 2 + 4 − 2 2 ) + ( 2 − 4 − 2 2 ) x \times (x^{2} - 3 \sqrt{2}) = (2 + \sqrt{4 -2 \sqrt{2} } ) + (2 - \sqrt{4 -2 \sqrt{2} } )

x 3 − 3 2 x = 4 x^{3} - 3 \sqrt{2} x = 4

x 3 − 3 2 x + 4 = 8 x^{3} - 3 \sqrt {2} x + 4 = \boxed{8}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

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