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Chapter 21: Chapter 21

Problem 628

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Find the solutions of the equation \(\mathrm{x}^{3}-1=0\)

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The solutions for the equation \(x^3 - 1 = 0\) are: \(x = 1, \frac{-1 \pm \sqrt{-3}}{2}\).
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Rewrite the equation using a sum of cubes formula

First, we can recognize that the equation \(x^3 - 1 = 0\) can be written as a sum of cubes equation, where \(x^3 = 1\). Therefore, we have: \[x^3 - 1^3 = 0\]

Step 2: Use the sum of cubes formula to factor the equation

The sum of cubes formula states that \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). However, since we have a difference of cubes, we need to slightly modify the formula to \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Now, substitute \(a = x\) and \(b = 1\) into the difference of cubes formula: \[(x - 1)(x^2 + x + 1) = 0\]

Step 3: Find the solutions for the equation

To find the solutions of the equation, we need to find all the possible values for \(x\) that satisfy the equation. To do this, we need to set each factor equal to zero. For the first factor, we have: \[x - 1 = 0\] \[x = 1\] For the second factor, we have a quadratic equation: \[x^2 + x + 1 = 0\] We can find the roots of the quadratic equation using the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where the coefficients are \(a = 1\), \(b = 1\), and \(c = 1\).

Step 4: Apply the quadratic formula

Using the coefficients and the quadratic formula, we get: \[x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}\] \[x = \frac{-1 \pm \sqrt{-3}}{2}\] Since the square root of a negative number is imaginary, the solutions for the quadratic equation are complex numbers. So, the solutions for the equation \(x^3 - 1 = 0\) are: \[x = 1, \frac{-1 \pm \sqrt{-3}}{2}\]

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