Solving the Cube Edge and Volume Increase Problem

This question asks us to find the original length of the edge of a cube, given that if the edge is increased by 4 cm, the volume increases by 988 cm³.

Let's break down the problem:

• We are dealing with a cube, a three-dimensional shape where all edges are equal in length.
• The volume of a cube is calculated by cubing the length of its edge.
• We have a scenario involving an original cube and a new cube with a longer edge.
• The difference in their volumes is given.

Setting up the Equations for Cube Volume

Let the original length of each edge of the cube be \(a\) cm.

The original volume of the cube, \(V_{original}\), is given by the formula:

\(V_{original} = a^3\)

When the edge of the cube is increased by 4 cm, the new edge length becomes \((a+4)\) cm.

The new volume of the cube, \(V_{new}\), is:

\(V_{new} = (a+4)^3\)

According to the problem, the volume increases by 988 cm³. This means the difference between the new volume and the original volume is 988 cm³.

So, we can write the equation:

\(V_{new} - V_{original} = 988\)

Substituting the expressions for \(V_{new}\) and \(V_{original}\):

\((a+4)^3 - a^3 = 988\)

Solving the Algebraic Equation

Now we need to solve the equation \((a+4)^3 - a^3 = 988\) for \(a\).

First, let's expand the term \((a+4)^3\) using the cubic expansion formula: \((x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\).

Here, \(x=a\) and \(y=4\).

\((a+4)^3 = a^3 + 3(a^2)(4) + 3(a)(4^2) + 4^3\)

\((a+4)^3 = a^3 + 12a^2 + 3a(16) + 64\)

\((a+4)^3 = a^3 + 12a^2 + 48a + 64\)

Now substitute this expanded form back into our main equation:

\((a^3 + 12a^2 + 48a + 64) - a^3 = 988\)

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

\(12a^2 + 48a + 64 = 988\)

To solve for \(a\), rearrange the equation into a standard quadratic form \(Ax^2 + Bx + C = 0\) by subtracting 988 from both sides:

\(12a^2 + 48a + 64 - 988 = 0\)

\(12a^2 + 48a - 924 = 0\)

We can simplify this quadratic equation by dividing all terms by the greatest common divisor, which is 12:

\(\frac{12a^2}{12} + \frac{48a}{12} - \frac{924}{12} = \frac{0}{12}\)

\(a^2 + 4a - 77 = 0\)

Now we solve the quadratic equation \(a^2 + 4a - 77 = 0\). We can try to factor this quadratic. We need two numbers that multiply to -77 and add up to 4.

Factors of -77 are (1, -77), (-1, 77), (7, -11), (-7, 11).

The pair (-7, 11) adds up to 4 (-7 + 11 = 4) and multiplies to -77 (-7 * 11 = -77).

So, we can factor the quadratic as:

\((a - 7)(a + 11) = 0\)

This equation gives us two possible solutions for \(a\):

\(a - 7 = 0 \Rightarrow a = 7\)

\(a + 11 = 0 \Rightarrow a = -11\)

Since \(a\) represents the length of the edge of a cube, it must be a positive value. Therefore, we discard the solution \(a = -11\).

The original length of each edge of the cube is \(a = 7\) cm.

Verification of the Original Edge Length

Let's check if an original edge length of 7 cm results in a volume increase of 988 cm³ when increased by 4 cm.

• Original edge length: 7 cm
• Original volume: \(V_{original} = 7^3 = 7 \times 7 \times 7 = 343\) cm³
• New edge length: \(7 + 4 = 11\) cm
• New volume: \(V_{new} = 11^3 = 11 \times 11 \times 11 = 1331\) cm³
• Volume increase: \(V_{new} - V_{original} = 1331 - 343 = 988\) cm³

The calculated volume increase matches the given information in the problem. Therefore, the original length of each edge of the cube is indeed 7 cm.

Revision Table: Cube Volume Basics

Additional Information: Solving Quadratic Equations

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

There are several methods to solve quadratic equations:

• Factoring: If the quadratic expression \(ax^2 + bx + c\) can be factored into the product of two linear factors, say \((px + q)(rx + s)\), then the solutions (roots) are found by setting each factor to zero: \(px + q = 0\) and \(rx + s = 0\). In our problem, \(a^2 + 4a - 77 = 0\) was factored as \((a-7)(a+11)=0\).
• Using the Quadratic Formula: The solutions to \(ax^2 + bx + c = 0\) are given by the formula:

  \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

  For \(a^2 + 4a - 77 = 0\), \(a=1\), \(b=4\), \(c=-77\). \(a = \frac{-4 \pm \sqrt{4^2 - 4(1)(-77)}}{2(1)}\) \(a = \frac{-4 \pm \sqrt{16 + 308}}{2}\) \(a = \frac{-4 \pm \sqrt{324}}{2}\) \(a = \frac{-4 \pm 18}{2}\) Two solutions: \(a = \frac{-4 + 18}{2} = \frac{14}{2} = 7\) and \(a = \frac{-4 - 18}{2} = \frac{-22}{2} = -11\).
• Completing the Square: Another method, though often more involved for general equations, can also be used.

In geometry problems involving lengths or physical quantities, we usually only consider the positive solutions obtained from solving such equations.