Cube Roots

Cube Roots exercise

• Complete the description of cube roots.

  Hints

  When finding the cube root, do we look for a number that adds to or multiplies by itself? How many times?

  This is the square root symbol ${\sqrt{}}$. What do we add to make it the cube root symbol?

  Look at the image, we use this method to help us find the cube root. What is the method called?

  Solution

  A cube root is a number that, when multiplied by itself three times, equals a given number.

  The mathematical symbol for the cube root of a number looks like the mathematical symbol for the square root of a number with an additional 3. It looks like this, ${\bf{\sqrt[3]{ }}}$.

  We use prime factorization to help us find the cube root of a number.

• Identify the cube root of a number

  Hints

  A cube root is a number that, when multiplied by itself three times, equals a given number.

  We can use prime factorization to find the cube root. In the image we can see how that is used to find the cube root of $125$.

  Cubes and cube roots are inverse operations (opposites). $5$ cubed is equal to $125$, therefore the cube root of $125$ is ...

  Solution

  $27 = 3 \times 3 \times 3$ so the cube root of $\bf{27}$ is $\bf{3}$.
  $125 = 5 \times 5 \times 5$ so the cube root of $\bf{125}$ is $\bf{5}$.
  $8 = 2 \times 2 \times 2$ so the cube root of $\bf{8}$ is $\bf{2}$.
  $1 = 1 \times 1 \times 1$ so the cube root of $\bf{1}$ is $\bf{1}$.

• Identify the cube roots

  Hints

  We can use prime factorization to find the cube root of a number. For example, if we were finding the cube root of $27$ then $27 = 3 \times 3 \times 3$.

  Cubes and cube roots are inverse operations (opposites) if you know that $3^3=27$ then you know that the cube root of $27$ is $3$.

  Solution

  • $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 = (2 \times 5)(2 \times 5)(2 \times 5) = 10 \times 10 \times 10 = 10^3$ so, $\bf{\sqrt[3]{1,000} =10}$.

  • $343= 7 \times 7 \times 7 = 7^3$ so, $\bf{\sqrt[3]{343} =7}$.

  • $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2 \times 3) \times (2 \times 3) \times (2 \times 3) = 6 \times 6 \times 6 = 6^3$ so, $\bf{\sqrt[3]{216} =6}$.
  • $729 = 9 \times 9 \times 9 = 9^3$ so, $\bf{\sqrt[3]{729} =9}$.

• Use prime factorization to find the cube root of a larger number.

  Hints

  Find the prime factorization - this is a large number so you will want to notice that even numbers can be divided by $2$ first.

  Do you know your divisibility rules? For example if a number can be divided by $3$ the sum of the digits is divisible by $3$.

  $2 + 7 + 4 + 4 = 17$, $17$ is not divisible by $3$ so $2{,}744$ is not divisible by $3$ either.

  $2{,}744 = 2 \times 2 \times 2 \times 7 \times 7 \times 7$

  Solution

  $2{,}744 = 2 \times 2 \times 2 \times 7 \times 7 \times 7 = 2 \times 7 \times 2 \times 7 \times 2 \times 7 = (2 \times 7) \times (2 \times 7) \times (2 \times 7) = 14 \times 14 \times 14 = 14^3$

  Therefore $\bf{\sqrt[3]{2{,}744}}$ = $14$.

• What is the $\sqrt[3]{64}$?

  Hints

  What is the prime factorization of $64$?

  $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$

  We need three groups of numbers so we can rearrange this to show
  $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) = 4 \times 4 \times 4$

  Solution

  $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$
  $2 \times 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) = 4 \times 4 \times 4$
  $4 \times 4 \times 4 = 4^3$
  The $\sqrt[\bf{3}]{\bf{64}}$ is $\bf{4}$.

• Identify numbers that are a perfect cube.

  Hints

  A perfect cube root will have a solution that is an integer (whole number).

  $300 = 2 \times 2 \times 3 \times 5 \times 5$, this cannot be rearranged into three identical groups so $\bf{300}$ does not have a perfect cube root.

  Another way we know that $300$ does not have a perfect cube root is by looking at the perfect cubes either side of it.
  $6 \times 6 \times 6 = 216$
  $7 \times 7 \times 7 = 343$
  $300$ is between $216$ and $343$ so the cube root of $\bf{300}$ is greater than $\bf{6}$ but less than $\bf{7}$ . It cannot be an integer answer and therefore does not have a perfect cube root.

  Solution

  $\bf{343}$, $\bf{1{,}331}$, and $\bf{3{,}375}$ are perfect cubes.

  • $343 = 7 \times 7 \times 7$ so $343$ has a perfect cube root.
  • $1{,}331 = 11 \times 11 \times 11$ so $1{,}331$ has a perfect cube root.
  • $3{,}375 = 15 \times 15 \times 15$ so $3{,}375$ has a perfect cube root.

  $\bf{3{,}00}$, $\bf{3{,}000}$ and $\bf{3{,}300}$ are not perfect cubes.

  • $300 = 2 \times 2 \times 3 \times 5 \times 5$ so $300$ does not have a perfect cube root.
  • $3{,}300 = 33 \times 100 = 2 \times 2 \times 3 \times 5 \times 5 \times 11$ so $3{,}300$ does not have a perfect cube root.
  • $3{,}000 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 \times 5$ so $3{,}000$ does not have a perfect cube root.