Correct. $\displaystyle{27}{{y}}^{{2}}={3}\cdot{3}\cdot{3}\cdot{y}\cdot{y}\usepackage{cmbright}\Huge\$ and $\displaystyle{18}{{y}}^{{3}}={2}\cdot{3}\cdot{3}\cdot{y}\cdot{y}\cdot{y}\usepackage{cmbright}\Huge\$ , so the LCD $\displaystyle={2}\cdot{3}\cdot{3}\cdot{3}\cdot{y}\cdot{y}\cdot{y}={54}{{y}}^{{3}}\usepackage{cmbright}\Huge\$ .

$\displaystyle\frac{{{4}{x}}}{{{27}{{y}}^{{2}}}}+\frac{{{5}{x}}}{{{18}{{y}}^{{3}}}}=\frac{{{\left({4}{x}\right)}{\left({2}{y}\right)}}}{{{\left({27}{{y}}^{{2}}\right)}{\left({2}{y}\right)}}}+\frac{{{\left({5}{x}\right)}{\left({3}\right)}}}{{{\left({18}{{y}}^{{3}}\right)}{\left({3}\right)}}}\usepackage{cmbright}\Huge\$

$\displaystyle=\frac{{{8}{x}{y}+{15}{x}}}{{{54}{{y}}^{{3}}}}=\frac{{{x}{\left({8}{y}+{15}\right)}}}{{{54}{{y}}^{{3}}}}\usepackage{cmbright}\Huge\$