Inspired by Swapnil Das

Algebra Level 3

Rationalize the denominator in

$\frac{ 1 } { \sqrt{2} + \sqrt[3] {3 } }.$

2 - \sqrt[3]{3}

\frac{ \sqrt[3] { 3 } - \sqrt{2} } { \sqrt[3]{9} - 2 }

4\sqrt[3]{3}-3\sqrt{2}\sqrt[3]{3}+3\sqrt[3]{9}-2\sqrt{2}\sqrt[3]{9}-4\sqrt{2}+6

\sqrt{2} + \sqrt[3] { 3} - 2

2 solutions

Vighnesh Raut
Mar 24, 2015

$\displaystyle{\frac { 1 }{ \sqrt { 2 } +\sqrt [ 3 ]{ 3 } } =\frac { 1 }{ \sqrt { 2 } +\sqrt [ 3 ]{ 3 } } \left( \frac { \sqrt { 2 } -\sqrt [ 3 ]{ 3 } }{ \sqrt { 2 } -\sqrt [ 3 ]{ 3 } } \right) }$ $\displaystyle{=\frac { \sqrt { 2 } -\sqrt [ 3 ]{ 3 } }{ 2-{ \left( \sqrt [ 3 ]{ 3 } \right) }^{ 2 } } \\ =\frac { \sqrt [ 3 ]{ 3 } -\sqrt { 2 } }{ \sqrt [ 3 ]{ 9 } -2 } }$ We know the identity $\displaystyle{{ a }^{ 3 }-{ b }^{ 3 }=(a-b)({ a }^{ 2 }+ab+{ b }^{ 2 })}$ So using this, we multiply the numerator and denominator with ${ \sqrt [ 3 ]{ 9 } }^{ 2 }+2\sqrt [ 3 ]{ 9 } +{ 2 }^{ 2 }$ . Thus, $\displaystyle{\frac { \sqrt [ 3 ]{ 3 } -\sqrt { 2 } }{ \sqrt [ 3 ]{ 9 } -2 } \\ =\left( \frac { \sqrt [ 3 ]{ 3 } -\sqrt { 2 } }{ \sqrt [ 3 ]{ 9 } -2 } \right) \left( \frac { { \sqrt [ 3 ]{ 9 } }^{ 2 }+2\sqrt [ 3 ]{ 9 } +{ 2 }^{ 2 } }{ { \sqrt [ 3 ]{ 9 } }^{ 2 }+2\sqrt [ 3 ]{ 9 } +{ 2 }^{ 2 } } \right) \\ =\frac { \left( \sqrt [ 3 ]{ 3 } -\sqrt { 2 } \right) \left( { \sqrt [ 3 ]{ 9 } }^{ 2 }+2\sqrt [ 3 ]{ 9 } +{ 2 }^{ 2 } \right) }{ 9-8 } \\ =4\sqrt [ 3 ]{ 3 } -3\sqrt { 2 } \sqrt [ 3 ]{ 3 } +3\sqrt [ 3 ]{ 9 } -2\sqrt { 2 } \sqrt [ 3 ]{ 3 } -4\sqrt { 2 } +6\\ \\ }$

Good approach, dealing with the square root first and then the cube root. Is there another way to figure out what the process should be like?

Calvin Lin Staff - 6 years, 2 months ago

$x^6 - y^6 = (x+y)(x^5 - x^4 y + x^3 y^2 - x^2 y^3 + xy^4 - y^5)$

Pi Han Goh - 6 years, 2 months ago

That's one way of thinking about it.

More generally, we look for the minimal polynomial. For $x = \sqrt{2} + \sqrt[3]{3}$ , we have $x^6 - 6x^4 - 6x^3 + 12x^2 - 36 x + 1 = 0$ . This gives us $\frac{1}{x} = - ( x^5 - 6x^3 - 6x^2 + 12x - 36)$ , which is the answer that we want.

The minimial polynomial can be found by taking $( x - \sqrt{2} ) ^ 3 -3 = 0$ which gives us $x^3 - 3 \sqrt{2} x^2 + 6x - 2 \sqrt{2} - 3 = 0$ , and then taking $( x^3 + 6x -3)^2 = ( 3 \sqrt{2} x^2 + 2\sqrt{2})^2$ , or that $x^6 + 12x^4 - 6x^3 + 36x^2 - 36x + 9 = 18x^4 + 24x^2 + 8$ . Upon simplification, we get the above equation.

Calvin Lin Staff - 6 years, 2 months ago

@Calvin Lin – Thanks for more general method.

Niranjan Khanderia - 6 years, 2 months ago

@Calvin Lin – well i didn't understand it

can anyone make me understand??

For your information I am 15 yrs old

Ashwin Upadhyay - 6 years, 2 months ago

yeah, I also think that the point is to make the exponent of both the term same. Then trying to rationalize the denominator with standard procedures or identities.

Soumo Mukherjee - 6 years, 2 months ago

Not a very different approach from the above one,

$\displaystyle \cfrac { 1 }{ \sqrt { 2 } +\sqrt [ 3 ]{ 3 } } =\cfrac { 1 }{ { \left( { 8 }^{ { 1 }/{ 2 } } \right) }^{ { 1 }/{ 3 } }+{ \left( { 9 }^{ { 1 }/{ 2 } } \right) }^{ { 1 }/{ 3 } } }$

Then multiplying with ${ 8 }^{ { 1 }/{ 3 } }+{ 9 }^{ { 1 }/{ 3 } }-{ \left( 8\cdot 9 \right) }^{ { 1 }/{ 6 } }$ . To get rid of the cube root. Then dealt with the square root.

Still digging for other approaches :)

Soumo Mukherjee - 6 years, 2 months ago

Niranjan Khanderia
Mar 27, 2015

We may first use the identity $a^3+b^3 =(a+b)(a^2-ab+b^2) ~and ~then ~~ (a+b)(a-b)=a^2-b^2,$ that makes calculation a little simpler. Going to sixth power , though is a general method needs a little more calculations. Approach in two steps, divides the problem in to two simple parts.

Inspired by Swapnil Das

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