Solve Equations With Cube Root \( \sqrt[3]{x} \)

Tutorial on how to solve equations containing cube roots. Detailed solutions, explanations, and exercises are included.

The idea behind solving equations containing cube roots is to raise both sides to the power 3 in order to clear the cube root using the property:

\[ (\sqrt[3]{x})^3 = x \]

Examples with Solutions

Example 1: Find all real solutions to \( \sqrt[3]{x} - x = 0 \)

Solution to Example 1

  • Rewrite the equation with the cube root isolated. \[ \sqrt[3]{x} = x \]
  • Raise both sides to the power 3. \[ (\sqrt[3]{x})^3 = x^3 \]
  • Rewrite with right-hand side equal to zero. \[ x - x^3 = 0 \]
  • Factor. \[ x(1 - x^2) = 0 \]
  • Solve for \(x\). Solutions: \(x = 0,\; x = -1,\; x = 1\).

Check the solutions

1. \(x = 0\)

LS: \(\sqrt[3]{0} - 0 = 0\)    RS: \(0\)

2. \(x = -1\)

LS: \(\sqrt[3]{-1} - (-1) = -1 + 1 = 0\)    RS: \(0\)

3. \(x = 1\)

LS: \(\sqrt[3]{1} - 1 = 0\)    RS: \(0\)

Example 2: Find all real solutions to \( \sqrt[3]{x^2 + 2x + 8} = 2 \)

Solution to Example 2

  • Given: \[ \sqrt[3]{x^2 + 2x + 8} = 2 \]
  • Raise both sides to the power 3. \[ (\sqrt[3]{x^2 + 2x + 8})^3 = 2^3 \]
  • Simplify. \[ x^2 + 2x + 8 = 8 \]
  • Rewrite with right-hand side equal to zero. \[ x^2 + 2x = 0 \]
  • Factor. \[ x(x + 2) = 0 \]
  • Solve. \[ x = 0,\; x = -2 \]

Checking as an exercise

1. \(x = 0\)   LS: \(\sqrt[3]{8} = 2\)   RS: \(2\)

2. \(x = -2\)   LS: \(\sqrt[3]{8} = 2\)   RS: \(2\)

Exercises

Solve the following equations:

  1. \( \sqrt[3]{x} - 4x = 0 \)
  2. \( \sqrt[3]{x^2 + 2x + 61} = 4 \)
  3. \( \sqrt[3]{x-2} + \sqrt[3]{x-1} = \sqrt[3]{2x-3} \) Challenging

Solutions

1. \( x = 0,\; x = \dfrac{1}{8},\; x = -\dfrac{1}{8} \)
2. \( x = 1,\; x = -3 \)
3. \( x = 1,\; x = \dfrac{3}{2},\; x = 2 \)

Detailed Solution for Exercise 3

Given the equation:

\[ \sqrt[3]{x-2} + \sqrt[3]{x-1} = \sqrt[3]{2x-3} \]

Cube both sides using the identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\):

\[ (x-2) + (x-1) + 3\sqrt[3]{(x-2)(x-1)}(\sqrt[3]{x-2} + \sqrt[3]{x-1}) = 2x - 3 \]

Substitute the original expression \(\sqrt[3]{x-2} + \sqrt[3]{x-1} = \sqrt[3]{2x-3}\) into the middle term:

\[ 2x - 3 + 3\sqrt[3]{(x-2)(x-1)(2x-3)} = 2x - 3 \]

Subtract \(2x - 3\) from both sides:

\[ 3\sqrt[3]{(x-2)(x-1)(2x-3)} = 0 \]

Dividing by 3 and cubing both sides yields:

\[ (x-2)(x-1)(2x-3) = 0 \]

Solving for \(x\) gives the complete set of solutions:

\[ x = 1, \; x = \frac{3}{2}, \; x = 2 \]