The cube roots of 125 are 5, and the two complex numbers $-\frac{5}{2} + \frac{5i\sqrt{3}}{2}$ and $-\frac{5}{2} - \frac{5i\sqrt{3}}{2}$.
A cube root of a number is a value that, when multiplied by itself three times, yields the original number. Every number, whether real or complex, has exactly three cube roots in the complex number system. One of these is a real number (the principal cube root), and the other two are complex conjugates.
Understanding Cube Roots
When we speak of "the cube root" of a number, we often refer to the principal (real) cube root. For example, the cube root of 8 is 2, because $2 \times 2 \times 2 = 8$. However, when the question asks for "all the cube roots," it includes the complex roots as well.
Finding the Principal (Real) Cube Root of 125
The real cube root of 125 is the positive real number that, when cubed, equals 125. We can find this by using prime factorization:
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Prime Factorization: Break down 125 into its prime factors.
$125 = 5 \times 25$
$125 = 5 \times 5 \times 5$ -
Grouping Factors: To find the cube root, we look for groups of three identical prime factors. In this case, we have three 5s.
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Result: Since $5 \times 5 \times 5 = 125$, the principal (real) cube root of 125 is 5.
We can write this as $\sqrt[3]{125} = 5$.
For more details on finding cube roots using prime factorization, you can refer to resources on finding cube roots.
Finding the Complex Cube Roots of 125
To find all cube roots of 125, we need to solve the equation $x^3 = 125$.
This can be rewritten as a polynomial equation:
$x^3 - 125 = 0$
We can factor this using the difference of cubes formula, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Here, $a=x$ and $b=5$:
$(x - 5)(x^2 + 5x + 25) = 0$
From this factored form, we can find the three roots:
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First Root (Real):
Set the first factor to zero:
$x - 5 = 0$
$x = 5$
This is our principal (real) cube root. -
Second and Third Roots (Complex):
Set the second factor to zero:
$x^2 + 5x + 25 = 0$
This is a quadratic equation, which can be solved using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, $a=1$, $b=5$, and $c=25$.$x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times 25}}{2 \times 1}$
$x = \frac{-5 \pm \sqrt{25 - 100}}{2}$
$x = \frac{-5 \pm \sqrt{-75}}{2}$Since $\sqrt{-75} = \sqrt{25 \times -3} = \sqrt{25} \times \sqrt{-3} = 5i\sqrt{3}$:
$x = \frac{-5 \pm 5i\sqrt{3}}{2}$
This gives us the two complex cube roots:
- $x_2 = -\frac{5}{2} + \frac{5i\sqrt{3}}{2}$
- $x_3 = -\frac{5}{2} - \frac{5i\sqrt{3}}{2}$
Summary of All Cube Roots of 125
The three cube roots of 125 are:
Type of Root | Value |
---|---|
Real Root | $5$ |
Complex Root 1 | $-\frac{5}{2} + \frac{5i\sqrt{3}}{2}$ |
Complex Root 2 | $-\frac{5}{2} - \frac{5i\sqrt{3}}{2}$ |