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Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.

What I have tried: Put $x = \frac{1}{9}$. We have, $\sqrt[3]3\big(4x^\frac{1}{3} - 2x^\frac{1}{3} + x^\frac{1}{3})$ $$\rightarrow \sqrt[3]3\big(x^\frac{1}{3}(4-2+1)\big)$$ $$\rightarrow \sqrt[3]3 \times \frac{1}{3\sqrt[3]x}$$ $$\rightarrow \frac{3}{27x} = \frac{1}{9x}$$ Putting back the value of $x$ gives the answer to be $1$.

However, in my text the answer is given to be $\sqrt[3]2 + 1$ , so where did I go wrong?

Can anyone help me?

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    $\begingroup$ Your first line is wrong how did you get $4x^{(1/3)}+2x^{1/3}$ $\endgroup$ Jan 2, 2021 at 8:37
  • $\begingroup$ Why? Because we are taking the cube root. $\endgroup$
    – Anonymous
    Jan 2, 2021 at 8:39
  • $\begingroup$ i fail to see how $$\sqrt[3]{4/9}=4x^{1/3}$$ $\endgroup$ Jan 2, 2021 at 8:40
  • $\begingroup$ Oh, I think the mistake went to $4x^\frac{1}{3}$ instead of $(4x)^\frac{1}{3}$. $\endgroup$
    – Anonymous
    Jan 2, 2021 at 8:41

1 Answer 1

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: $$\sqrt[3]{4/9}-\sqrt[3]{2/9}+\sqrt[3]{1/9}=\frac{(2^{1/3}+1)(2^{2/3}-2^{1/3}+1)}{(2^{1/3}+1)(3^{2/3})}=\frac{3}{3^{2/3}(2^{1/3}+1)}$$ so:$$\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}=\sqrt[3]{3}\cdot \frac{(2^{1/3}+1)(3^{2/3})}{3}=\sqrt[3]{2}+1$$

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  • $\begingroup$ Ok I am getting it, then do we cube the fraction? $\endgroup$
    – Anonymous
    Jan 2, 2021 at 8:54
  • $\begingroup$ @Anonymous no,no take the reciprocal and multiply with $\sqrt[3]{3}$ $\endgroup$ Jan 2, 2021 at 8:55
  • $\begingroup$ Ok, then I suppose we will get $\frac{√3}{\sqrt[3]2 + 1}$. $\endgroup$
    – Anonymous
    Jan 2, 2021 at 8:56
  • $\begingroup$ @Anonymous edited $\endgroup$ Jan 2, 2021 at 9:05
  • $\begingroup$ Oh nevermind, I feel a little stupid sometimes, I got it. $\endgroup$
    – Anonymous
    Jan 2, 2021 at 9:08

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