you would expand becausee it isn't equal to anything
(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one
so
for the 3rd power
\((a+b)^3=1a^3+3a^2b^1+3a^1b^2+1b^3\)
a=x
b=3
\((x+3)^3=1x^3+3x^2(3^1)+3x^1(3)^2+1(3)^3\)=
\(x^3+9x^2+27x+27\)
