Cube and cube roots grade VIII

Math Problems and Solutions - Cubes and Roots

1. Question: Find the smallest number by which the following must be divided to obtain a perfect cube.

   (a) 13718

   (b) 28672

   Solution:

   (a) Prime factorization of 13718 is \(2 \times 3 \times 3 \times 761\). To make it a perfect cube, it must be divided by \(2 \times 3 = 6\).

   (b) Prime factorization of 28672 is \(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\). To make it a perfect cube, it must be divided by \(2 \times 2 \times 3 = 12\).

2. Question: Find the volume of a cubical box whose surface area is \(486 \, cm^2\).

   Solution: Let the side of the cubical box be \(a\). The surface area of the cubical box is \(6a^2\). We are given that \(6a^2 = 486\), so \(a^2 = \frac{486}{6} = 81\). Therefore, \(a = 9\). The volume of the cubical box is \(a^3 = 9^3 = 729 \, cm^3\).

3. Question: Find the cube of \(-\frac{36}{7}\).

   Solution: The cube of \(-\frac{36}{7}\) is \(\left(-\frac{36}{7}\right)^3 = -\frac{36^3}{7^3} = -\frac{46656}{343} \approx -136.123\).

4. Question: Divide the number 8748 by the smallest number so that the quotient is a perfect cube. Also, find the cube root of the quotient.

   Solution: Prime factorization of 8748 is \(2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\). To make it a perfect cube, it must be divided by \(2 \times 2 \times 3 \times 3 = 12\). The quotient is \(\frac{8748}{12} = 729\), which is a perfect cube. The cube root of 729 is \(\sqrt[3]{729} = 9\).