To solve the problem, we need to determine the length of the edge of a new cube formed by melting three cubes with side lengths \(6 \, \text{cm}\), \(8 \, \text{cm}\), and \(1 \, \text{cm}\). The volume of the new cube will be equal to the sum of the volumes of the three original cubes.
Step 1: Calculate the volume of each original cube.
The volume \(V\) of a cube with side length \(a\) is given by:
\[ V = a^3 \]
For the first cube (\(a = 6 \, \text{cm}\)):
\[ V_1 = 6^3 = 216 \, \text{cm}^3 \]
For the second cube (\(a = 8 \, \text{cm}\)):
\[ V_2 = 8^3 = 512 \, \text{cm}^3 \]
For the third cube (\(a = 1 \, \text{cm}\)):
\[ V_3 = 1^3 = 1 \, \text{cm}^3 \]
Step 2: Find the total volume of the three cubes.
The total volume \(V_{\text{total}}\) is the sum of the volumes of the three cubes:
\[ V_{\text{total}} = V_1 + V_2 + V_3 = 216 + 512 + 1 = 729 \, \text{cm}^3 \]
Step 3: Determine the side length of the new cube.
Let the side length of the new cube be \(a_{\text{new}}\). The volume of the new cube is also given by \(a_{\text{new}}^3\). Since the volume of the new cube is equal to the total volume of the three original cubes:
\[ a_{\text{new}}^3 = 729 \]
Solve for \(a_{\text{new}}\) by taking the cube root of both sides:
\[ a_{\text{new}} = \sqrt[3]{729} = 9 \, \text{cm} \]
Final Answer:
The length of the edge of the new cube is \(9 \, \text{cm}\).
\[ {1} \]
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
If \( S \) and \( S' \) are the foci of the ellipse \[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \] and \( P \) is a point on the ellipse, then \[ \min (SP \cdot S'P) + \max (SP \cdot S'P) \] is equal to: