Question

tim bought 128 sandbags to completely fill a cube - shaped sandbox. each bag fills a cubic foot in the sandbox. what is the length, in feet, of one of the sides of the sandbox?

Explanation:

Step1: Find the volume of the sandbox

Since each bag fills \(1\) cubic foot and there are \(128\) bags, the volume \(V\) of the cube - shaped sandbox is \(V = 128\) cubic feet.

Step2: Use the formula for the volume of a cube

The formula for the volume of a cube is \(V=s^{3}\), where \(s\) is the side - length of the cube. So we have the equation \(s^{3}=128\).

Step3: Solve for \(s\)

We can rewrite \(128\) as \(128 = 64\times2=2^{6}\times2 = 2^{7}\). Also, \(s^{3}=128\), so \(s=\sqrt[3]{128}\). Since \(128 = 64\times2\), \(\sqrt[3]{128}=\sqrt[3]{64\times2}\). Using the property \(\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}\) (\(a = 64\), \(b = 2\)), and \(\sqrt[3]{64}=4\) (because \(4\times4\times4 = 64\)), we get \(s = 4\sqrt[3]{2}\approx4\times1.26=5.04\). But if we consider the prime - factorization \(128=2^{7}\), and using the rule \(\sqrt[3]{2^{7}}=2^{\frac{7}{3}}=2^{2+\frac{1}{3}}=4\times2^{\frac{1}{3}}\). Another way: \(s^{3}=128\), and \(128 = 2\times2\times2\times2\times2\times2\times2\). Group the factors in triples: \(s^{3}=2^{3}\times2^{3}\times2\). So \(s = 2\times2\times\sqrt[3]{2}=4\sqrt[3]{2}\approx5.04\).

Answer:

The length of one of the sides of the sandbox is \(4\sqrt[3]{2}\approx5.04\) feet.