Problem 37 - Intermediate Algebra

Question

Solve each equation. See Example 5. $$ \sqrt[3]{x^{3}-7}=x-1 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 2\) and \(x = -1\).

Step 1: Cubing Both Sides

First, eliminate the cube root by cubing both sides of the equation: \[( oot{3}{x^3 - 7})^3 = (x - 1)^3\]which simplifies to:\[x^3 - 7 = (x - 1)^3\]

Step 2: Expand the Cubic Binomial

Expand the right side of the equation, which involves a binomial raised to the third power:\[(x - 1)^3 = (x - 1)(x - 1)(x - 1) = x^3 - 3x^2 + 3x - 1\]

Step 3: Equate Both Expressions

Now we equate the two expressions:\[x^3 - 7 = x^3 - 3x^2 + 3x - 1\]

Step 4: Simplify the Equation

Subtract \(x^3\) from both sides to simplify:\[-7 = -3x^2 + 3x - 1\]Then add 1 to both sides to obtain:\[-6 = -3x^2 + 3x\]

Step 5: Solve the Quadratic Equation

Multiply the entire equation by -1 to make it positive:\[6 = 3x^2 - 3x\]Divide the entire equation by 3 to further simplify:\[2 = x^2 - x\]Rearrange to form a standard quadratic equation:\[x^2 - x - 2 = 0\]Factor the quadratic:\[(x - 2)(x + 1) = 0\]which gives solutions for \(x\):\[x = 2\quad \text{or} \quad x = -1\]

Step 6: Verify the Solutions

Substitute \(x = 2\) back into the original equation to check:\[\root{3}{2^3 - 7} = 2 - 1\]\[\root{3}{1} = 1\quad \text{True}\]Now check \(x = -1\):\[\root{3}{(-1)^3 - 7} = -1 - 1\]\[\root{3}{-8} = -2\quad \text{True}\]Both solutions satisfy the original equation.

Key Concepts

Cube Roots

Understanding cube roots is essential for solving equations like \(\sqrt[3]{x^3 - 7} = x - 1\). A cube root of a number \(a\) is another number \(b\) such that \(b^3 = a\). In our problem, the cube root is applied to the expression \(x^3 - 7\), which means we need to find a number that, when cubed, will give us this expression. Cube roots are crucial when dealing with radicals, particularly in reducing cube-based expressions. They allow us to clear the radical by cubing both sides of the equation, which is what was done immediately by cubing \(\sqrt[3]{x^3 - 7}\) to simplify it to \(x^3 - 7\).
By understanding cube roots, you will be better equipped to tackle similar problems where cube roots are involved either in the original equations or during steps to isolate variables.

Quadratic Equations

Quadratic equations appear frequently in algebra, especially when solving problems like the one given. A standard quadratic equation has the form \(ax^2 + bx + c = 0\). In our situation, after some algebraic manipulation, we ended up with \(x^2 - x - 2 = 0\). Quadratics can be solved in multiple ways, including factoring, using the quadratic formula, or completing the square.
For this specific problem, factoring was used, where the quadratic \(x^2 - x - 2\) was broken into two binomial factors: \((x - 2)(x + 1) = 0\). This provides an excellent example of how knowing the roots of a quadratic can directly give us the solutions to the equation. Identifying these solutions is crucial, and verifying them by substituting back into the original equation ensures their correctness.

Binomial Expansion

Binomial expansion is necessary when dealing with expressions raised to a power, such as \((x - 1)^3\) in our problem. The binomial theorem provides a systematic way to expand expressions of the form \((a + b)^n\). For a binomial cube, like \((x - 1)^3\), it requires expanding the product \((x - 1)(x - 1)(x - 1)\).
By distributing each term, we systematically break it down using the FOIL method first on two binomials \((x - 1)(x - 1)\), and then multiplying the result by the last binomial again. This expansion gives us \(x^3 - 3x^2 + 3x - 1\), showing how each term in the expansion contributes to the final polynomial.

• First, compute \((x - 1)(x - 1)\) to obtain \(x^2 - 2x + 1\).
• Then multiply that result by \((x - 1)\) again.
• Continue this process until the expression is fully expanded.

Mastering binomial expansions not only aids in simplifying equations but also prepares you for tackling more complex polynomials.