Problem 75 - Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ n=\left(n^{3}+n^{2}-1\right)^{1 / 3} $$

Step-by-Step Solution

Answer

The solutions are \(n = 1\) and \(n = -1\). Both are valid.

Step 1: Understand the problem

We need to solve the equation \(n = (n^3 + n^2 - 1)^{1/3}\). This involves finding values of \(n\) that satisfy the equation.

Step 2: Eliminate the cube root

Cube both sides of the equation to eliminate the cube root. We have: \(n^3 = n^3 + n^2 - 1\).

Step 3: Simplify the equation

Subtract \(n^3\) from both sides of the equation: \(0 = n^2 - 1\).

Step 4: Solve the simplified equation

Factor the equation: \(n^2 - 1 = (n-1)(n+1) = 0\). Hence, the solutions are \(n = 1\) and \(n = -1\).

Step 5: Verify solutions for extraneous roots

Substitute \(n = 1\) back into the original equation: \(1 = (1^3 + 1^2 - 1)^{1/3} = 1\). This is valid. Substitute \(n = -1\) into the original equation: \(-1 = ((-1)^3 + (-1)^2 - 1)^{1/3} = (-1)^{1/3}\). Evaluate the cube root: \((-1)^{1/3} = -1\). This solution is also valid.

Key Concepts

Cube Roots

Cube roots can be a bit tricky, but they're important in equations like the one we're dealing with here. The cube root of a number \(x\), noted as \(x^{1/3}\), is a value \(y\) such that \(y^3 = x\). This means raising \(y\) to the power of three will give you \(x\).

• For positive values, the cube root is straightforward, like the cube root of 8 is 2, because \(2^3 = 8\).
• For negative numbers, the cube root is also negative, because multiplying three negative numbers gives a negative result. For instance, the cube root of -8 is -2, as \((-2)^3 = -8\).

The beauty of cube roots in equations is that they allow us to solve for unknowns more effectively by leveling off an equation by cubing both sides. This moves us from a root-based equation to a polynomial one, which is often simpler to solve.

Extraneous Solutions

Extraneous solutions are potential solutions that appear in the process of solving an equation, but don't satisfy the original equation. They often come from operations like squaring or cubing both sides, which can introduce additional solutions due to the nature of these operations.

• When you cube both sides of an equation to eliminate a cube root, be aware that the new equation might have more solutions than the original.
• That's why it's crucial to substitute solutions back into the original equation to check their validity.

In our exercise, we substituted our solutions back into the original equation to ensure they worked. Both \(n = 1\) and \(n = -1\) checked out, meaning we had no extraneous solutions here. But always make this step a habit to avoid mistakes!

Factoring Equations

Factoring is a key part of solving polynomial equations, and it often simplifies these equations into more manageable pieces. In our given equation, after simplifying, we arrived at the polynomial \(n^2 - 1 = 0\). This can be factored as \((n-1)(n+1) = 0\).

• The rule here is that if a product of terms equals zero, then at least one of the terms must be zero.
• This leads us to set each factor equal to zero: \(n - 1 = 0\) and \(n + 1 = 0\), resulting in solutions \(n = 1\) and \(n = -1\).

Factoring isn't just a nifty shortcut; it often guides us to the solutions with less effort. Remember to look for common patterns like difference of squares or simple groupings to make equations simpler and more intuitive to solve.