Question
If a and b are distinct primes such that $\sqrt[3]{\text{a}^6\text{b}^{-4}}=\text{a}^\text{x}\text{b}^{2\text{y}},$find x and y.
✓
Answer
$\sqrt[3]{\text{a}^6\text{b}^{-4}}=\text{a}^\text{x}\text{b}^{2\text{y}}$$\Rightarrow(\text{a}^6\text{b}^{-4})^\frac{1}{3}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow\text{a}^{6\times\frac{1}{3}}\times\text{b}^{-\frac{4}{3}}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow\text{a}^2\times\text{b}^{-\frac{4}{3}}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow2=\text{x}$ and $-\frac{4}{3}=2\text{y}$
$\Rightarrow\text{x}=2$ and $\text{y}=-\frac{2}{3}$
$\Rightarrow\text{a}^{6\times\frac{1}{3}}\times\text{b}^{-\frac{4}{3}}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow\text{a}^2\times\text{b}^{-\frac{4}{3}}=\text{a}^\text{x}\times\text{b}^{2\text{y}}$
$\Rightarrow2=\text{x}$ and $-\frac{4}{3}=2\text{y}$
$\Rightarrow\text{x}=2$ and $\text{y}=-\frac{2}{3}$
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