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Problem 628

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

Expert verified
The solutions for the equation $$x^3 - 1 = 0$$ are: $$x = 1, \frac{-1 \pm \sqrt{-3}}{2}$$.
See the step by step solution

## 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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