Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$

Question

Given that

$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$ find the value of rationals $a,b,c.$

Solution I tried: I tried to rationalize it, but I'm not getting the answer:

$$\frac{1}{(1+\sqrt[3]{2})}\times \frac{(1-\sqrt[3]{2})}{(1-\sqrt[3]{2})}$$

$$\frac{(1-\sqrt[3]{2})}{1-2^{\frac{2}{3}}}$$ so doing so not getting answer; also, I tried to expand it, but we have condition that $(1+x)^n$ where $n$ is in fraction can be expandable only when $x < 1$, but here the cube root of $2$ is not less than $1.$

Thank you

Answer 1

Hint:

Rationalize the denominator by multiplying top and bottom by $1-\sqrt[3]2+\sqrt[3]4$

Answer 2

$\alpha=1+\sqrt[3]{2}$ is a root of $(x-1)^3=2$, and so $$ 3 = \alpha^3 - 3 \alpha^2 + 3 \alpha = \alpha(\alpha^2 - 3 \alpha + 3) $$ Therefore, $$ \alpha^{-1}=\frac13(\alpha^2 - 3 \alpha + 3) $$ Now use that $$ \alpha^2 = 1+2\sqrt[3]2+\sqrt[3]4 $$

This solution works in general because every nonzero algebraic number is a root of a polynomial with nonzero independent term.

Answer 3

If you multiply both sides by $1+\sqrt[3]2$ you get $$1=(1+\sqrt[3]2)(a+b\sqrt[3]2+c\sqrt[3]4)\\ =a+2c+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$$ which we can resolve into three equations $$1=a+2c\\0=a+b\\0=b+c$$ by equating the parts proportional to $1,\sqrt[3]2,\sqrt[3]4$ which give $$b=-\frac 13\\ a=\frac 13\\c=\frac 13$$

Answer 4

So we have $\frac 1{1 + \sqrt[3]2}$ and we want to get

$\frac 1{1 + \sqrt[3]2}\frac {something}{something}=\frac{something}{something\ with\ no\ radicals}$

Taking a page from doing this for square roots where, from $a + \sqrt b$ we realize $(a+\sqrt b)(a-\sqrt b) = a^2 - b$, which works because $(m-n)(m+n) = m^2 -n^2$.

If we use the idea $(m\pm k)(m^{n-1} \mp m^{n-2}k + ......) = m^n \pm k^n$.

So if we can see $(1+\sqrt[3]{2})(1 - \sqrt[3]{2} + \sqrt[3]{2}^2) = 1 + \sqrt[3]2^3 = 1+2=3$

And $\frac 1{1 + \sqrt[3]2}=\frac 1{1 + \sqrt[3]2}\frac {1-\sqrt[3]2 + \sqrt[3]2^2} {1-\sqrt[3]2 + \sqrt[3]2^2}= \frac {1-\sqrt[3]2 + \sqrt[3]2^2}3=\frac 13 -\frac 13\sqrt[3]2 + \frac 13\sqrt[3]{2^2}$

Or $a = c =\frac 13$ and $b =-\frac 13$

Answer 5

Utilize $$\frac1{a+1}=\frac{a^2-a+1}{a^3+1}$$

with $a = \sqrt[3]{2}$ to obtain

$$\frac1{1+\sqrt[3]{2}}= \frac{(\sqrt[3]{2})^2-\sqrt[3]{2}+1}{(\sqrt[3]{2})^3+1 }=\frac13(\sqrt[3]{2^2}-\sqrt[3]{2}+1) $$