PreálgebraRoots and Irrational Numbers8.EE.A.2

Cube Roots

∛64 = 4 because 4³ = 64.

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Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.

Cubing uses a number as a factor three times: $4^{3} = 4 \cdot 4 \cdot 4 = 64$. A cube root undoes cubing, the way a square root undoes squaring.

Por qué funciona

A cube with edges 4 units long is built from 4 layers of $4 \times 4$ unit cubes: $4 \cdot 4 \cdot 4 = 64$ cubes. So if a cube holds 64 unit cubes, its edge is 4. We write $\sqrt[3]{64} = 4$.
A $4 \times 4 \times 4$ cube holds 64 unit cubes. Drag to turn it, and try other sizes.

Idea clave

$\sqrt[3]{n}$ is the number whose cube is $n$. The perfect cubes are $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, \ldots$

Unlike square roots, cube roots of negative numbers exist. Three negative factors make a negative, so $(-4)^{3} = -64$ and $\sqrt[3]{-64} = -4$.

Worked examples

Ejemplo resuelto: A perfect cube

Find $\sqrt[3]{125}$.
  1. Which number, used three times as a factor, gives 125?
  2. $5 \cdot 5 \cdot 5 = 125$.

Respuesta: $5$

Ejemplo resuelto: A negative number

Find $\sqrt[3]{-27}$.
  1. $3^{3} = 27$, so try $-3$.
  2. $(-3)(-3)(-3) = 9 \cdot (-3) = -27$.

Respuesta: $-3$

Ejemplo resuelto: A fraction

Find $\sqrt[3]{\dfrac{8}{27}}$.
  1. Take the cube root of the top and of the bottom: $\sqrt[3]{8} = 2$ and $\sqrt[3]{27} = 3$.
  2. Check: $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{27}$.

Respuesta: $\frac{2}{3}$

Most numbers are not perfect cubes. Their cube roots still have a place between whole numbers: $3^{3} = 27$ and $4^{3} = 64$, so $\sqrt[3]{50}$ is between 3 and 4.

Error común

Dividing by 3, as in $\sqrt[3]{27} = 9$. Check by cubing: $9^{3} = 729$. The cube root of 27 is 3, because $3 \cdot 3 \cdot 3 = 27$.

Práctica

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Fuentes y lecturas

  • OpenStax Intermediate Algebra 2e, Section 8.1: Simplify Expressions with Roots
  • Common Core State Standards for Mathematics, Grade 8, 8.EE.A.2