Solving System of Equations with Cubes

We are given a system of two equations involving variables \(a\) and \(b\):

1. \(a^3 = 335 + b^3\)
2. \(a = 5 + b\)

We are also given the conditions that \(a > 0\) and \(b > 0\).

Our goal is to find the value of \(a + b\).

Step 1: Rearrange the first equation

Let's move the \(b^3\) term from the right side to the left side in the first equation:

\(a^3 - b^3 = 335\)

Step 2: Apply the difference of cubes identity

Recall the algebraic identity for the difference of cubes: \(x^3 - y^3 = (x - y)(x^2 + xy + y^2)\).

Applying this identity to \(a^3 - b^3\), we get:

\((a - b)(a^2 + ab + b^2) = 335\)

Step 3: Use the second equation to find (a - b)

The second given equation is \(a = 5 + b\).

We can rearrange this equation to find the value of \(a - b\):

\(a - b = 5\)

Step 4: Substitute (a - b) into the difference of cubes equation

Now, substitute the value of \(a - b = 5\) into the equation from Step 2:

\(5(a^2 + ab + b^2) = 335\)

Step 5: Solve for the expression involving squares

Divide both sides of the equation by 5:

\(a^2 + ab + b^2 = \frac{335}{5}\)

\(a^2 + ab + b^2 = 67\)

Step 6: Substitute 'a' from the second equation into the new equation

We have the equation \(a^2 + ab + b^2 = 67\) and the relationship \(a = 5 + b\).

Substitute \((5 + b)\) for \(a\) in the equation \(a^2 + ab + b^2 = 67\):

\((5 + b)^2 + (5 + b)b + b^2 = 67\)

Step 7: Expand and simplify the equation

Expand the terms:

• \((5 + b)^2 = 5^2 + 2(5)(b) + b^2 = 25 + 10b + b^2\)
• \((5 + b)b = 5b + b^2\)

Substitute these back into the equation:

\((25 + 10b + b^2) + (5b + b^2) + b^2 = 67\)

Combine like terms:

\(3b^2 + 15b + 25 = 67\)

Step 8: Form a quadratic equation

Subtract 67 from both sides to set the equation to zero:

\(3b^2 + 15b + 25 - 67 = 0\)

\(3b^2 + 15b - 42 = 0\)

Step 9: Solve the quadratic equation for 'b'

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

\(b^2 + 5b - 14 = 0\)

Now, we can solve this quadratic equation. We can factor it by finding two numbers that multiply to -14 and add up to 5. These numbers are 7 and -2.

So, the equation can be factored as:

\((b + 7)(b - 2) = 0\)

This gives two possible solutions for \(b\):

• \(b + 7 = 0 \implies b = -7\)
• \(b - 2 = 0 \implies b = 2\)

Step 10: Use the condition b > 0 to find the correct value of 'b'

The problem states that \(b > 0\). Therefore, we discard the solution \(b = -7\) and choose \(b = 2\).

Step 11: Find the value of 'a'

Now that we have the value of \(b\), we can find the value of \(a\) using the second equation \(a = 5 + b\):

\(a = 5 + 2\)

\(a = 7\)

We check if \(a > 0\). Yes, \(7 > 0\), so this solution is valid.

Step 12: Calculate a + b

Finally, we need to find the value of \(a + b\):

\(a + b = 7 + 2\)

\(a + b = 9\)

Verification

Let's quickly check if these values of \(a=7\) and \(b=2\) satisfy the original equations:

Equation 1: \(a^3 = 335 + b^3\)

\(7^3 = 343\)

\(335 + 2^3 = 335 + 8 = 343\)

\(343 = 343\). This equation holds true.

Equation 2: \(a = 5 + b\)

\(7 = 5 + 2\)

\(7 = 7\). This equation also holds true.

The conditions \(a > 0\) and \(b > 0\) are also met since \(a=7\) and \(b=2\).

Thus, the value of \(a + b\) is 9.

Revision Table: Key Concepts

Additional Information: Algebraic Identities and Solving Equations

Understanding algebraic identities is fundamental for simplifying and solving many mathematical problems, including those involving powers like cubes. The difference of cubes identity used in this problem is one example. Others include:

• Sum of Cubes: \(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\)
• Difference of Squares: \(x^2 - y^2 = (x - y)(x + y)\)
• Perfect Square Trinomials: \((x+y)^2 = x^2 + 2xy + y^2\), \((x-y)^2 = x^2 - 2xy + y^2\)

When solving systems of equations, the substitution method, as demonstrated here, is a powerful technique, especially when one equation provides a simple expression for one variable in terms of the other. Elimination is another common method.

Solving quadratic equations is a frequent step in algebra. Besides factoring, you can also use the quadratic formula (\(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)) or completing the square to find solutions for \(Ax^2 + Bx + C = 0\).

Always pay close attention to any conditions given for the variables, such as being positive, negative, or integers, as these conditions help in selecting the valid solutions from the possibilities obtained during the solving process.