Question

investigate roots

1. a puzzle maker built the cube puzzle shown.

a. what is the edge length of the cube puzzle? explain.
b. suppose the puzzle maker wants to build a larger cube puzzle and wants the volume to be a perfect cube when measured in cubic inches. if the volume must be less than 1000 cubic inches, what are some volumes the puzzle maker could use? explain.
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Explanation:

Part A

Step1: Assume volume of cube

From the handwritten note, the volume seems to be \(216\) cubic units (since \(6^3 = 216\)). For a cube, volume \(V = s^3\), where \(s\) is edge length.

Step2: Solve for edge length

To find \(s\), take the cube root of volume: \(s=\sqrt[3]{V}\). If \(V = 216\), then \(s=\sqrt[3]{216}\). Since \(6\times6\times6 = 216\), \(\sqrt[3]{216}=6\).

Step1: Recall perfect cube formula

A perfect cube has volume \(V = n^3\), where \(n\) is a positive integer, and \(V<1000\).

Step2: Find valid \(n\) values

We need \(n^3<1000\). Solve for \(n\): \(n<\sqrt[3]{1000}=10\). So \(n\) can be \(1,2,3,4,5,6,7,8,9\) (excluding \(n = 10\) as \(10^3=1000\) which is not less than \(1000\)).

Step3: Calculate volumes

• For \(n = 1\): \(V = 1^3=1\)
• For \(n = 2\): \(V = 2^3 = 8\)
• For \(n = 3\): \(V = 3^3=27\)
• For \(n = 4\): \(V = 4^3 = 64\)
• For \(n = 5\): \(V = 5^3=125\)
• For \(n = 6\): \(V = 6^3 = 216\) (already used in part A, but still valid)
• For \(n = 7\): \(V = 7^3=343\)
• For \(n = 8\): \(V = 8^3 = 512\)
• For \(n = 9\): \(V = 9^3=729\)

These volumes are perfect cubes (since they are cubes of integers) and less than \(1000\) cubic inches.

Answer:

The edge length of the cube puzzle is \(6\) units (assuming the volume is \(216\) cubic units) because the volume of a cube is \(s^3\), and taking the cube root of \(216\) (since \(6^3 = 216\)) gives the edge length \(s = 6\).

Part B