## Algebra 1

$0$
Given : $(3x^{3}y^{a})^{3}=27x^{9}$ (The empty square box is represented as $a$) Hence, $3^{3}$ X $x^{9}$ X $y^{3a}=27x^{9}$ (since $(ab)^{n} = a^{n}b^{n}$ and $(a^{m})^{n}=a^{mn})$ This becomes : $27$ X $x^{9}$ X $y^{3a}$ = $27$ X $x^{9}$ X $y^{0}$ ($3^{3} = 27$ and $y^{0} = 1$) This implies : $y^{3a}$ = $y^{0}$ Thus, $3a=0$ and $a=\frac{0}{3}=0$ Final result : $0$