`(x^2 + y^2)^4` Use the Binomial Theorem to expand and simplify the expression.

You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace `x^2` for x, `y^2` for y and 4 for n, such that:

`(x^2+y^2)^4 = 4C0 (x^2)^4 + 4C1 (x^2)^3*(y^2)^1 + 4C2 (x^2)^2*(y^2)^2 +4C3 (x^2)*(y^2)^3 +4C4 (y^2)^4`

By definition, nC0 = nCn = 1, hence `4C0 = 4C4 = 1` .

By definition nC1 = nC(n-1) = n, hence `4C1 = 4C3 = 4.`

By definition `nC2 = nC(n-2) = (n(n-1))/2` , hence `4C2 = (4(4-1))/2 = 6`

`(x^2+y^2)^4 = x^8 + 4(x^6)*(y^2) +6 (x^4)*(y^4) +4 (x^2)*(y^6)+ (y^8)`

Hence, expanding the complex number using binomial theorem yields the simplified result `(x^2+y^2)^4 = x^8 + 4(x^6)*(y^2) +6 (x^4)*(y^4) +4 (x^2)*(y^6)+ (y^8).`