Answer:
[tex](x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5[/tex]
[tex](x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6[/tex]
Step-by-step explanation:
In order to find the values of [tex](x+y)^5[/tex] and [tex](x+y)^6[/tex], you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).
[tex](x+y)^5[/tex] = [tex]\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i[/tex] = [tex]\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5[/tex] = [tex]x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5[/tex].
[tex](x+y)^6[/tex] = [tex]\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i[/tex] = [tex]\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6[/tex]= [tex]x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6[/tex].
