## Precalculus (6th Edition) Blitzer

$x^3+9x^2y+27xy^2+27y^3$
Binomial Theorem or Binomial expansion can be defined as: $(x+y)^n=\displaystyle \binom{n}{0}x^ny^0+\displaystyle \binom{n}{1}x^{n-1}y^1+........+\displaystyle \binom{n}{n}x^0y^n$ Need to apply the formula to get the Binomial Expansion. we have $(x+3y)^3=\displaystyle \binom{3}{0}(x)^3(3y)^0+\displaystyle \binom{3}{1}(x^{2})(3y)^1 +\displaystyle \binom{3}{2}(x)^1(3y)^2+\displaystyle \binom{3}{3}(x)^0(3y)^3$ $=x^3(1)+(3)(x^2)(3y)+3(x)(9y^2)+27(y^3)(1)$ $\bf{(Simplify)}$ $=x^3+9x^2y+27xy^2+27y^3$