Answer
$(x^2+y^2)^5=\binom{5}{0}(x^2)^5(y^2)^0+\binom{5}{1}(x^2)^4(y^2)^1+\binom{5}{2}(x^2)^3(y^2)^2+\binom{5}{3}(x^2)^2(y^2)^3+\binom{5}{4}(x^2)^1(y^2)^4+\binom{5}{5}(x^2)^0(y^2)^5=x^{10}\times 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}\times 1=$
$x^{10} 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}$
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 $x^2$ and y becomes $y^2$; $n=5$
$(x^2+y^2)^5=\binom{5}{0}(x^2)^5(y^2)^0+\binom{5}{1}(x^2)^4(y^2)^1+\binom{5}{2}(x^2)^3(y^2)^2+\binom{5}{3}(x^2)^2(y^2)^3+\binom{5}{4}(x^2)^1(y^2)^4+\binom{5}{5}(x^2)^0(y^2)^5=x^{10}\times 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}\times 1=$
$x^{10} 1+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}$
