Understanding the Algebraic Problem

The question provides us with two algebraic equations involving two positive variables, \(a\) and \(b\). We are given:

• Equation 1: \(a^3 = 117 + b^3\)
• Equation 2: \(a = 3 + b\)

We are also told that \(a > 0\) and \(b > 0\). Our goal is to find the value of \(a + b\).

Rewriting the Equations

Let's rearrange Equation 1 to group the terms involving \(a\) and \(b\):

From \(a^3 = 117 + b^3\), we can subtract \(b^3\) from both sides to get:

\[a^3 - b^3 = 117\]

Equation 2 gives us a direct relationship between \(a\) and \(b\):

\[a = 3 + b\]

This can also be written as \(a - b = 3\).

Applying the Difference of Cubes Identity

The expression \(a^3 - b^3\) on the left side of the rewritten Equation 1 is a difference of cubes. We can use the algebraic identity for the difference of cubes, which is:

\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]

Substituting this into the rewritten Equation 1 (\(a^3 - b^3 = 117\)), we get:

\[(a - b)(a^2 + ab + b^2) = 117\]

Using the Relationship a - b = 3

From Equation 2, we found that \(a - b = 3\). We can substitute this value into the equation above:

\[(3)(a^2 + ab + b^2) = 117\]

Now, divide both sides by 3:

\[a^2 + ab + b^2 = \frac{117}{3}\]

\[a^2 + ab + b^2 = 39\]

We now have a new equation relating \(a\) and \(b\): \(a^2 + ab + b^2 = 39\).

Finding the Values of a and b

We have the equation \(a^2 + ab + b^2 = 39\) and the relationship \(a = 3 + b\) from the original equations. We can substitute the expression for \(a\) from \(a = 3 + b\) into the new equation \(a^2 + ab + b^2 = 39\) to solve for \(b\).

Substitute \(a = 3 + b\):

\[(3 + b)^2 + (3 + b)b + b^2 = 39\]

Expand the terms:

• \((3 + b)^2 = 3^2 + 2(3)(b) + b^2 = 9 + 6b + b^2\)
• \((3 + b)b = 3b + b^2\)

Substitute these expanded terms back into the equation:

\[(9 + 6b + b^2) + (3b + b^2) + b^2 = 39\]

Combine like terms:

• \(b^2 + b^2 + b^2 = 3b^2\)
• \(6b + 3b = 9b\)
• \(9\) (constant term)

The equation becomes:

\[3b^2 + 9b + 9 = 39\]

Subtract 39 from both sides to set the quadratic equation to zero:

\[3b^2 + 9b + 9 - 39 = 0\]

\[3b^2 + 9b - 30 = 0\]

We can simplify this quadratic equation by dividing the entire equation by 3:

\[\frac{3b^2}{3} + \frac{9b}{3} - \frac{30}{3} = \frac{0}{3}\]

\[b^2 + 3b - 10 = 0\]

Now we need to solve this quadratic equation for \(b\). We can factor the quadratic expression \(b^2 + 3b - 10\). We look for two numbers that multiply to -10 and add up to 3. These numbers are 5 and -2.

So, we can factor the equation as:

\[(b + 5)(b - 2) = 0\]

This gives us two possible values for \(b\):

• \(b + 5 = 0 \implies b = -5\)
• \(b - 2 = 0 \implies b = 2\)

The problem states that \(b > 0\). Therefore, we must choose the positive value for \(b\).

So, \(b = 2\).

Now that we have the value of \(b\), we can find the value of \(a\) using the relationship \(a = 3 + b\):

\[a = 3 + 2\]

\[a = 5\]

We check if \(a > 0\). Yes, \(5 > 0\).

Calculating a + b

We have found that \(a = 5\) and \(b = 2\). The question asks for the value of \(a + b\).

\[a + b = 5 + 2\]

\[a + b = 7\]

Verification

Let's check if these values satisfy the original equations:

• Equation 1: \(a^3 = 117 + b^3\)
• Substitute \(a=5\) and \(b=2\): \(5^3 = 117 + 2^3\)
• \(125 = 117 + 8\)
• \(125 = 125\) (This equation holds true)

• Equation 2: \(a = 3 + b\)
• Substitute \(a=5\) and \(b=2\): \(5 = 3 + 2\)
• \(5 = 5\) (This equation also holds true)

Both original equations are satisfied by \(a=5\) and \(b=2\), and the conditions \(a > 0\) and \(b > 0\) are met. The value of \(a + b\) is 7.

Revision Table: Key Steps in Solving the Algebraic Problem

Additional Information: Algebraic Identities and Solving Quadratics

This problem effectively uses a common algebraic identity and the method for solving quadratic equations. Understanding these concepts is crucial for solving many algebra problems.

Difference of Cubes Identity

The identity used is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). There is also a sum of cubes identity: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). These identities are very useful for factoring expressions involving cubes.

Solving Quadratic Equations

A quadratic equation is an equation of the form \(Ax^2 + Bx + C = 0\), where \(x\) is the variable and \(A\), \(B\), and \(C\) are constants with \(A \neq 0\). In our case, the variable was \(b\), and the equation was \(b^2 + 3b - 10 = 0\). Here, \(A=1\), \(B=3\), and \(C=-10\). Quadratic equations can be solved by:

• Factoring (as we did in this solution)
• Using the quadratic formula: \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)
• Completing the square

Factoring is often the quickest method when it is possible.