Beginning Algebra With Applications, Chapter 7, 7.2, Section 7.2, Problem 88

Simplify $\displaystyle 9x^3 (3x^2 y)^2 - x(x^3 y)^2$


$\begin{equation} \begin{aligned} 9x^3 (3x^2 y)^2 - x(x^3 y)^2 =& 9x^3 (3)^2 x^4 y^2 - x \left( x^6 y^2 \right) && \text{Multiply each exponent in $3x^2 y$ and in $x^3 y$ by the exponent outside the parentheses} \\ \\ =& (9x^3) \left( 9x^4 y^2 \right) - x \left( x^6 y^2 \right) && \text{Simplify } (3)^2 \\ \\ =& (9 \cdot 9) \left( x^3 \cdot x^4 \right) y^2 - (x \cdot x^6) y^2 && \text{Use Properties of Multiplication to rearrange and group factors} \\ \\ =& 81x^7 y^2 - x^7 y^2 && \text{Multiply variables with the same base by adding the exponents} \\ \\ =& 80 x^7 y^2 && \end{aligned} \end{equation}$