Answer
$(ax+by)^5=\binom{5}{0}(ax)^5(by)^0+\binom{5}{1}(ax)^4(by)^1+\binom{5}{2}(ax)^3(by)^2+\binom{5}{3}(ax)^2(by)^3+\binom{5}{4}(ax)^1(by)^4+\binom{5}{5}(ax)^0(by)^5=a^5x^5+5a^4bx^4y+10a^3b^2x^3y^2+10a^2b^3x^2y^3+5ab^4xy^4+b^5y^5$
Work Step by Step
The binomial theorem expands an algebraic expression in the form of: $(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{i}x^{n-i}y^i+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}x^0y^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^i$
Here, x becomes $ax$ , y becomes $by$, and $n=5$.
$(ax+by)^5=\binom{5}{0}(ax)^5(by)^0+\binom{5}{1}(ax)^4(by)^1+\binom{5}{2}(ax)^3(by)^2+\binom{5}{3}(ax)^2(by)^3+\binom{5}{4}(ax)^1(by)^4+\binom{5}{5}(ax)^0(by)^5=a^5x^5+5a^4bx^4y+10a^3b^2x^3y^2+10a^2b^3x^2y^3+5ab^4xy^4+b^5y^5$
