Algebra: A Combined Approach (4th Edition)

a.$x^{3}$ b. $81y^{16}$ c. $\frac{x^3}{64}$
a. When dividing $x^{7}$ by $x^{4}$, you simply subtract the exponents because the base number/variable is the same. So 7-4 equals 3. b.$(3y^{4})^4$ Simply multiply the exponent by the base number within the parentheses. 3 has an understood exponent of 1. y has an exponent of 4. $3^4$ equals 81. $(y^4)^4$ equals $y^{16}$. c. This is similar to the last problem. x gets cubed, so it's $x^3$. 4 gets cubed as well, which equals 64. It remains as a fraction and can't be reduced.