Simple but Nice #2

Algebra Level 3

Given that x = 2 + 2 2 / 3 + 2 1 / 3 x = 2 + 2^{2/3} + 2^{1/3} , what is

x 3 6 x 2 + 6 x ? x^3 - 6x^2 + 6x?


The answer is 2.

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7 solutions

Dhruva Patil
Jan 5, 2015

X = 2 + 2 2 3 + 2 1 3 X = 2 1 3 ( 2 2 3 + 2 1 3 + 1 ) X = 2 1 3 ( 2 2 3 + 2 1 3 + 2 2 + 1 ) X = 2 1 3 ( X 1 ) C u b i n g b o t h s i d e s X 3 = 2 ( X 3 1 3 X 2 + 3 X ) X 3 = 2 X 3 2 6 X 2 + 6 X X 3 6 X 2 + 6 X = 2 X=2+{ 2 }^{ \frac { 2 }{ 3 } }+{ 2 }^{ \frac { 1 }{ 3 } }\\ X={ 2 }^{ \frac { 1 }{ 3 } }({ 2 }^{ \frac { 2 }{ 3 } }+{ 2 }^{ \frac { 1 }{ 3 } }+1)\\ X={ 2 }^{ \frac { 1 }{ 3 } }({ 2 }^{ \frac { 2 }{ 3 } }+{ 2 }^{ \frac { 1 }{ 3 } }+2-2+1)\\ X={ 2 }^{ \frac { 1 }{ 3 } }(X-1)\\ Cubing\quad both\quad sides\\ { X }^{ 3 }=2({ X }^{ 3 }-1-3{ X }^{ 2 }+3X)\\ { X }^{ 3 }=2{ X }^{ 3 }-2-6{ X }^{ 2 }+6X\\ { X }^{ 3 }-6{ X }^{ 2 }+6X=\boxed{2}

Very clear.

Roman Frago - 6 years, 5 months ago

my reply is 2(1+1+1)=6 therefore x=6 given polynomial 6^3 - 6(6^2) + 6(6)=6^3 - 6^3 + 36 = 36 all your steps are inunderstandable and what about mine it is absolutely right

sudoku subbu - 6 years, 5 months ago

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X 2 ( 1 + 1 + 1 ) X = 2 1 3 ( 2 2 3 + 2 1 3 + 1 ) X \neq 2(1+1+1) \\ X={ 2 }^{ \frac { 1 }{ 3 } }({ 2 }^{ \frac { 2 }{ 3 } }+{ 2 }^{ \frac { 1 }{ 3 } }+1)

Dhruva Patil - 6 years, 5 months ago

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oh sorry i realized when i went to my school. i was in a sleepy mode while i was in brilliant

sudoku subbu - 6 years, 5 months ago
Sanjeet Raria
Jan 5, 2015

x 2 = 2 1 3 ( 2 1 3 + 1 ) x-2=2^{\frac{1}{3}}( 2^{\frac{1}{3}}+1) Cubing gives us, x 3 6 x 2 + 12 x 8 = 6 + 6 × 2 1 3 ( 2 1 3 + 1 ) x^3-6x^2+12x-8=6+6× 2^{\frac{1}{3}}( 2^{\frac{1}{3}}+1) = 6 + 6 ( x 2 ) =6+6(x-2) x 3 6 x 2 + 6 x = 2 \Rightarrow x^3-6x^2+6x=\boxed{2}

We are given that x = 2 + 2 3 + 4 3 x=2+\sqrt[3]{2}+\sqrt[3]{4} . Let x = y + 2 x=y+2 , hence y = 2 3 + 4 3 y=\sqrt[3]{2}+\sqrt[3]{4} .

Cube both sides, simplify and substitute again the value of y y :

y 3 = ( 2 3 + 4 3 ) 3 y 3 = 2 + 3 8 3 ( 2 3 + 4 3 ) + 4 y 3 = 6 y + 6 y 3 6 y 6 = 0 y^3=(\sqrt[3]{2}+\sqrt[3]{4})^3 \\y^3=2+3\sqrt[3]{8}(\sqrt[3]{2}+\sqrt[3]{4})+4 \\ y^3=6y+6 \\ y^3-6y-6=0

Finally, since x = y + 2 x=y+2 , we know that y = x 2 y=x-2 , so:

( x 2 ) 3 6 ( x 2 ) 6 = 0 x 3 6 x 2 + 12 x 8 6 x + 12 6 = 0 x 3 6 x 2 + 6 x = 2 (x-2)^3-6(x-2)-6=0 \\ x^3-6x^2+12x-8-6x+12-6=0 \\ x^3-6x^2+6x=\boxed{2}

This is exactly the same solution presented by Sanjeet Raria.

Roman Frago - 6 years, 5 months ago

x = 2 + 2 2 / 3 + 2 1 / 3 x=2+2^{2/3}+2^{1/3}

x 2 = 2 2 / 3 + 2 1 / 3 x-2=2^{2/3}+2^{1/3}

Cubing both sides:

( x 2 ) 3 = 6 + 6 × 2 2 / 3 + 6 × 2 1 / 3 = 6 x 6 (x-2)^3=6+6 \times 2^{2/3}+6 \times 2^{1/3} = 6x - 6

( x 2 ) 3 6 x + 8 = 2 (x-2)^3 - 6x + 8 = 2

x 3 6 x 2 + 6 x = 2 \boxed { x^3 - 6x^2 + 6x =2 }

Anna Anant
Jan 15, 2015

: x - 1 = 2^(-1/3)*x So (x-1)^3 = 0.5x^3 x^3 - 3x^2 + 3x - 1 = 0.5x^3 i.e. x^3 - 6x^2 + 6x = 2

Roman Frago
Jan 13, 2015

Doing it the long way. x 2 = 2 2 + ( 2 1 3 ) 2 + ( 2 2 3 ) 2 + 2 ( 2 ) ( 2 2 3 ) + 2 ( 2 ) ( 2 1 3 ) + 2 ( 2 1 3 ) ( 2 2 3 ) x^2=2^2+(2^{\frac {1} {3}})^2+(2^{\frac {2} {3}})^2+2(2)(2^{\frac {2} {3}})+2(2)(2^{\frac {1} {3}})+2(2^{\frac {1} {3}})(2^{\frac {2} {3}}) x 2 = 4 + 2 2 3 + 2 4 3 + 4 ( 2 2 3 ) + 4 ( 2 1 3 ) + 2 ( 2 ) x^2=4+2^{\frac {2} {3}}+2^{\frac {4} {3}}+ 4(2^{\frac {2} {3}})+ 4(2^{\frac {1} {3}})+2(2) x 2 = 8 + 5 ( 2 2 3 ) + 4 ( 2 1 3 ) + 2 4 3 x^2= 8+ 5(2^{\frac {2} {3}})+ 4(2^{\frac {1} {3}})+2^{\frac {4} {3}} x 2 = 8 + 5 ( 2 2 3 ) + 4 ( 2 1 3 ) + 2 ( 2 1 3 ) x^2= 8+ 5(2^{\frac {2} {3}})+ 4(2^{\frac {1} {3}})+2(2^{\frac {1} {3}}) x 2 = 8 + 5 ( 2 2 3 ) + 6 ( 2 1 3 ) x^2= 8+ 5(2^{\frac {2} {3}})+ 6(2^{\frac {1} {3}}) x 3 = x 2 × x x^3=x^2 \times x x 3 = [ 8 + 5 ( 2 2 3 ) + 6 ( 2 1 3 ) ] × [ 2 + 2 2 3 + 2 1 3 ) ] x^3= [8+ 5(2^{\frac {2} {3}})+ 6(2^{\frac {1} {3}})] \times [2+2^{\frac {2} {3}}+2^{\frac {1} {3}})] x 3 = 38 + 24 ( 2 2 3 ) + 30 ( 2 1 3 ) x^3=38+ 24(2^{\frac {2} {3}})+ 30(2^{\frac {1} {3}})

Thus x 3 6 x 2 + 6 x = 2 x^3-6x^2+6x=2

Richard Levine
Jan 11, 2015

Using a calculator: I can approximate x quickly as 2 + 1.587 + 1.26 = 4.8476 Plugging into x^3-6x^2+6x I get 113.914 - 140.995 + 29.0856 = 2.0046. I gave 2 as my answer. If it would have been wrong, then I would have gone the other route, performing the algebraic cube.

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