# (x+y)^5+(x-y)^3=?

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You should use binomial theorem to expand `(x+y)^5` and `(x-y)^3` such that:

`(x+y)^5 = C_5^0 x^5 + C_5^1 x^4y + C_5^2 x^3y^2 + C_5^3 x^2y^3 + C_5^4 xy^4 + C_5^5 y^5`

You need to know that `C_5^0 = C_5^5 = 1, C_5^1 = C_5^4 = 5 , C_5^2 = C_5^3 = (5(5-1))/2 = 10.`

Substituting the found values for binomial coefficients yields:

`(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5`

You need to raise to cube the binomial `x - y` such that:

`(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3`

`(x+y)^5 + (x-y)^3 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 + x^3 - 3x^2y + 3xy^2 - y^3`

**Hence, evaluating the sum of powers of binomials yields `(x+y)^5 + (x-y)^3 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 + x^3 - 3x^2y + 3xy^2 - y^3.` **