$\frac{1}{{x}^{3}}+\frac{1}{{y}^{7}}=?$

$\frac{1}{{x}^{3}{y}^{7}}$

$\frac{2}{{x}^{3}{y}^{7}}$

$\frac{{x}^{3}+{y}^{3}}{{x}^{3}{y}^{7}}$

$\frac{{x}^{7}+{y}^{7}}{{x}^{3}{y}^{7}}$

$\frac{{x}^{3}+{y}^{7}}{{x}^{3}{y}^{7}}$

**Solution:**

Since x^{3} is the highest (really, only) power of the base x, and since y^{7 }is the highest (only) power of the base y, the LCD of these two fractions is (x^{3})(y^{7}). Let’s convert each fraction to a denominator of (x^{3})(y^{7}) and combine the fractions:

$\begin{array}{l}\Rightarrow \frac{{y}^{7}}{{y}^{7}}\left(\frac{1}{{x}^{3}}\right)+\frac{{x}^{3}}{{x}^{3}}\left(\frac{1}{{y}^{7}}\right)\\ \Rightarrow \frac{{y}^{7}}{{x}^{3}{y}^{7}}+\frac{{x}^{3}}{{x}^{3}{y}^{7}}=\frac{{x}^{3}+{y}^{7}}{{x}^{3}{y}^{7}}\end{array}$