Q. 8.78 - Intermediate Algebra

Question

Simplify:

1.  $\sqrt{27p^3}-\sqrt{48p^3}$
2. $\sqrt[3]{256y^5}-\sqrt[3]{32n^5}$
   

Step-by-Step Solution

Answer

Part a. $-p\sqrt{3p}$

Part b. $4y\sqrt[3]{4y^2}-2n\sqrt[3]{4n^2}$

Part (a) Step 1. Simplify the radicals and combine like radicals

Using $\sqrt{ab}=\sqrt{a}·\sqrt{b}$ simplify the expression.

$\begin{array}{rcl}\sqrt{27p^3}-\sqrt{48p^3}& =& \sqrt{9p^2·3p}-\sqrt{16p^2·3p}\\ & =& \sqrt{9p^2}·\sqrt{3p}-\sqrt{16p^2}·\sqrt{3p}\\ & =& 3p\sqrt{3p}-4p\sqrt{3p}\end{array}$

On combing the like radicals we get

$\begin{array}{rcl}3p\sqrt{3p}-4p\sqrt{3p}& =& (3-4)p\sqrt{3p}\\ & =& -p\sqrt{3p}\end{array}$

Part (b) Step 1. Simplify the radicals and combine like radicals

Using $\sqrt{ab}=\sqrt{a}·\sqrt{b}$ simplify the expression.

$\begin{array}{rcl}\sqrt[3]{256y^5}-\sqrt[3]{32n^5}& =& \sqrt[3]{64y^3·4y^2}-\sqrt[3]{8n^3·4n^2}\\ & =& \sqrt[3]{64y^3}·\sqrt[3]{4y^2}-\sqrt[3]{8n^3}·\sqrt[3]{4n^2}\\ & =& 4y\sqrt[3]{4y^2}-2n\sqrt[3]{4n^2}\end{array}$

The indices are the same but the radicals are different. So they are not like radicals. And hence they cannot be simplified further.