Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises - Page 636: 11

$(x^{2}y-1)^{5}=x^{10}y^{5}-5x^{8}y^{4}+10x^{6}y^{3}-10x^{4}y^{2}+5x^{2}y-1$

We use the values from the 5th row of Pascal's Triangle to find the coefficients: $(x^{2}y-1)^{5}=(x^{2}y)^{5}-5(x^{2}y)^{4}+10(x^{2}y)^{3}-10(x^{2}y)^{2}+5(x^{2}y)^1-1=x^{10}y^{5}-5x^{8}y^{4}+10x^{6}y^{3}-10x^{4}y^{2}+5x^{2}y-1$