Solving the Cube Edge and Volume Increase Problem

This problem asks us to find the original edge length of a cube given information about how its volume changes when its edge is increased by a certain amount.

Understanding the Problem

A cube is a three-dimensional shape with six equal square faces. All its edges are of equal length. The volume of a cube is calculated by cubing the length of its edge.

• Let the original edge length of the cube be \(x\) cm.
• The original volume of the cube is \(V_{original} = x^3\) cm³.

The problem states that the edge of the cube is increased by 2 cm.

• The new edge length becomes \((x + 2)\) cm.
• The new volume of the cube is \(V_{new} = (x + 2)^3\) cm³.

We are also told that the volume increases by 488 cm³ when the edge is increased. This means the difference between the new volume and the original volume is 488 cm³.

So, we can write the equation:

Solving the Equation for the Edge Length

We need to expand the term \((x + 2)^3\). We can use the algebraic identity \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\).

Here, \(a = x\) and \(b = 2\). So, \((x + 2)^3\) expands to:

Now substitute this expansion back into our equation:

The \(x^3\) terms cancel out:

To solve for \(x\), we need to rearrange the equation into a standard quadratic form \(ax^2 + bx + c = 0\).

We can simplify this quadratic equation by dividing all terms by the common factor, which is 6:

Now we have a quadratic equation \(x^2 + 2x - 80 = 0\). We can solve this equation by factoring, completing the square, or using the quadratic formula. Factoring is often the simplest method if possible.

We need to find two numbers that multiply to -80 and add up to 2 (the coefficient of the \(x\) term). The pairs of factors for 80 are (1, 80), (2, 40), (4, 20), (5, 16), (8, 10). To get a product of -80 and a sum of +2, the numbers must have opposite signs, and the larger absolute value must be positive. The pair (10, -8) fits this criteria: \(10 \times (-8) = -80\) and \(10 + (-8) = 2\).

So, we can factor the quadratic equation as:

For this product to be zero, one or both of the factors must be zero.

• Case 1: \(x + 10 = 0 \Rightarrow x = -10\)
• Case 2: \(x - 8 = 0 \Rightarrow x = 8\)

The edge length of a cube must be a positive value. Therefore, the solution \(x = -10\) is not physically possible for a real cube.

The only valid solution for the original edge length is \(x = 8\) cm.

Verification

Let's check if an original edge length of 8 cm gives the stated increase in volume.

• Original edge = 8 cm
• Original volume = \(8^3 = 8 \times 8 \times 8 = 512\) cm³
• New edge = \(8 + 2 = 10\) cm
• New volume = \(10^3 = 10 \times 10 \times 10 = 1000\) cm³
• Increase in volume = \(V_{new} - V_{original} = 1000 - 512 = 488\) cm³

This matches the problem statement that the volume increases by 488 cm³. Therefore, the original edge length of the cube was 8 cm.

Final Answer

The length of each edge of the original cube is 8 cm.

Revision Table: Cube Volume Concepts

Additional Information: Algebraic Identities and Quadratic Equations

Solving problems like this often involves using algebraic identities and solving quadratic equations.

Algebraic Identity: Cube of a Sum

The identity used in this problem is the expansion of a binomial raised to the power of 3:

This identity is useful for expanding expressions like \((x + 2)^3\).

Quadratic Equations

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a \neq 0\).

There are several ways to solve quadratic equations:

• Factoring: If the quadratic expression \(ax^2 + bx + c\) can be factored into the form \((px + q)(rx + s)\), then the solutions (roots) are found by setting each factor to zero: \(px + q = 0\) and \(rx + s = 0\).
• Quadratic Formula: The solutions to the equation \(ax^2 + bx + c = 0\) are given by the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] This formula can be used for any quadratic equation.
• Completing the Square: This method involves rearranging the equation to form a perfect square trinomial on one side.

In our problem, the equation \(x^2 + 2x - 80 = 0\) is a quadratic equation where \(a = 1\), \(b = 2\), and \(c = -80\). We solved it by factoring.