Solução:
Pelo teorema binomial: ( x + 2 y ) 3 = ( 3 0 ) x 3 ( 2 y ) 0 + ( 3 1 ) x 2 ( 2 y ) 1 + ( 3 2 ) x 1 ( 2 y ) 2 + ( 3 3 ) x 0 ( 2 y ) 3 = x 3 + 6 x 2 y + 12 x y 2 + 8 y 3 {\displaystyle (x+2y)^{3}={\binom {3}{0}}x^{3}(2y)^{0}+{\binom {3}{1}}x^{2}(2y)^{1}+{\binom {3}{2}}x^{1}(2y)^{2}+{\binom {3}{3}}x^{0}(2y)^{3}=x^{3}+6x^{2}y+12xy^{2}+8y^{3}}
