Chapter 7, Problem 14CLR

To determine

Simplify the given equation.

Answer to Problem 14CLR

  $\boxed{\left(\frac{y^9}{y^3-64-12y^2+12by}\right)}$

Explanation of Solution

Formula used:

  $\begin{array}{l}{\left(a^m\right)}^n=a^{m\times n}\\ \\ {\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\end{array}$

Given information:

  ${\left(\frac{y^3}{y-4}\right)}^3$

  ${\left(\frac{y^3}{y-4}\right)}^3$

Now distributing exponent equally in numerator and denominator by using identity ${\left(a^m\right)}^n=a^{m\times n}$:

  $\begin{array}{l}\left(\frac{y^{3\times 3}}{{\left(y-4\right)}^{{}^{1\times 3}}}\right)\\ \\ =\left(\frac{y^9}{{\left(y-4\right)}^3}\right)\end{array}$

Now applying identity ${\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)$ on ${\left(y-4\right)}^3$

we get:

  $\begin{array}{l}=\left(\frac{y^9}{y^3-64-3\left(y\right)\left(4\right)\left(y-b\right)}\right)\\ \\ =\boxed{\left(\frac{y^9}{y^3-64-12y^2+12by}\right)}\end{array}$