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If $\mathrm{r}^{1 / 3}+\frac{1}{r^{1 / 3}}=3$ for a real number $\mathrm{r} \neq 0$, then what is
$\mathrm{r}+1$ equal to?
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$\mathrm{r}+1$ equal to?
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Verified Answer
The correct answer is:
18
Given equation is
$\mathrm{r}^{1 / 3}+\frac{1}{\mathrm{r}^{1 / 3}}=3$
Cubing both sides, we get
$\left(r^{1 / 3}+\frac{1}{r^{1 / 3}}\right)^{3}=3^{3}\left[(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$
$\Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+3\left(\mathrm{r}^{1 / 3}+\frac{1}{\mathrm{r}^{1 / 3}}\right)=27$
$\Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+3.3=27 \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+27-9=18$
$\mathrm{r}^{1 / 3}+\frac{1}{\mathrm{r}^{1 / 3}}=3$
Cubing both sides, we get
$\left(r^{1 / 3}+\frac{1}{r^{1 / 3}}\right)^{3}=3^{3}\left[(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$
$\Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+3\left(\mathrm{r}^{1 / 3}+\frac{1}{\mathrm{r}^{1 / 3}}\right)=27$
$\Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+3.3=27 \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}+27-9=18$
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