Simplifying radicals - resolved

[Xearf_987]

1/ (cbrt(2)-1)

How do you go about simplifying this? I know that if you have a problem like: 5/cbrt(2) .... you would multiply the problem by a clever form of one that, in this case, makes the denomonater a perfect cube. So you would multiply both the numerator and the denomonater by the cube root of 4 so that the denomonater becomes the cube root of 8 which is 2, thus eliminating the radical in the denomonater. I understand that if there's at least two terms, you would multiply the numerator and the denomonater by the conjugate of the denomonater. But what exactly do I do in this situation?

 

[soroban]

Re: Simplifying radicals

Hello, Xearf_987!

\(\displaystyle \L\frac{1}{\sqrt[3]{2}\,-\,1}\)

With a binomial denominator with a cube root, the procedure is trickier.


Recall that: \(\displaystyle \,(a\,-\,b)(a^2\,+\,ab\,+\,b^2)\:=\:a^3\,-\,b^3\)

\(\displaystyle \text{To "change" }\,a^{\frac{1}{3}}\,-\,b^{\frac{1}{3}}\, \text{ into: }\,\left(a^{\frac{1}{3}}\right)^3\,-\,\left(b^{\frac{1}{3}}\right)^3\;=\;a\,-\,b\)

\(\displaystyle \;\;\text{we must multiply top and bottom by: }\,\left(a^{\frac{2}{3}}\,+\,a^{\frac{1}{3}}b^{\frac{1}{3}}\,+\,b^{\frac{2}{3}}\right)\)


Your problem has: \(\displaystyle \,a\,=\,2,\;b\,=\,1\)

Multiply top and bottom by: \(\displaystyle \,\left(2^{\frac{2}{3}}\,+\,2^{\frac{1}{3}}\,+\,1\right)\,\) or \(\displaystyle \,\left(\sqrt[3]{4}\,+\,\sqrt[3]{2}\,+\,1\right)\)

 

[stapel]

You're right about needing a "clever" form. But cube roots do make things a little more difficult.

When you were rationalizing denominators with square roots, you multiplied by the conjugate, setting up a "difference of squares" thing that cancelled out the terms with the radicals.

In this case, you need to set up a "difference of cubes" thing.

. . . . .a³ - b³ = (a - b)(a² + ab + b²)

In this case, a = cbrt[2], b = 1, you have (a - b), and you need to multiply, top and bottom, by a² + ab + b²

No, it won't be pretty....

Eliz.

 

[Xearf_987]

Oh okay, I get it now. Thanks a lot!