Answer
$(x+3)^5=\binom{5}{0}x^53^0+\binom{5}{1}x^{4}3^1+\binom{5}{2}x^{3}3^2+\binom{5}{3}x^{2}3^3+\binom{5}{4}x^{1}3^{4}+\binom{5}{5}x^03^5=x^5+15x^4+90x^3+270x^2+405x+243$
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, $y=3; n=5$
$(x+3)^5=\binom{5}{0}x^53^0+\binom{5}{1}x^{4}3^1+\binom{5}{2}x^{3}3^2+\binom{5}{3}x^{2}3^3+\binom{5}{4}x^{1}3^{4}+\binom{5}{5}x^03^5=x^5+15x^4+90x^3+270x^2+405x+243$
