Question

Find the cube root of 1728 by prime factorisation method.

Answer 1

Hint: In this question, we need to find the cube root of 1728 by prime factorization method. For this, we will first find the prime factors of 1728. After that, we will pair similar factors in a group of 3 to be denoted as cubes. Those numbers (with cubes) will then be multiplied to get the required cube root.

Complete step by step answer:
Here we need to find the cube root of 1728 by the prime factorization method. For this, let us first find the prime factors of 1728.
Dividing 1728 by the smallest prime number 2, we obtain 864. Again dividing the obtained quotient 864 by 2 we get 432. Again dividing 432 by 2, we get 216. Dividing 216 by 2, we get 108. Dividing 108 by 2, we get 54. Dividing 54 by 2, we get 27. Now 27 cannot be divided by 2 so divide it by the next prime number which is 3, so we get 9. Dividing 9 by 3 we get 3. At last dividing 3 by 3, we get 1. Therefore, prime factorisation of 1728 is $ 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3 $ .
\[\begin{align}
  & 2\left| \!{\underline {\,
  1728 \,}} \right. \\
 & 2\left| \!{\underline {\,
  864 \,}} \right. \\
 & 2\left| \!{\underline {\,
  432 \,}} \right. \\
 & 2\left| \!{\underline {\,
  216 \,}} \right. \\
 & 2\left| \!{\underline {\,
  108 \,}} \right. \\
 & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 0\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Now let us pair the similar factors in the group of 3, we get, $ 1728=\left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 3\times 3\times 3 \right) $ .
We can represent these with the sign of cubes we get, $ 1728={{2}^{3}}\times {{2}^{3}}\times {{3}^{3}} $ .
Taking cube root on both sides we get, \[\sqrt[3]{1728}=\sqrt[3]{{{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}}\].
We know that \[\sqrt[3]{a\times b}=\sqrt[3]{a}\times \sqrt[3]{b}\] so we get, \[\sqrt[3]{1728}=\sqrt[3]{{{2}^{3}}}\times \sqrt[3]{{{2}^{3}}}\times \sqrt[3]{{{3}^{3}}}\].
Now we know that cube and cube root cancels each other. Therefore we get \[\sqrt[3]{1728}=2\times 2\times 3=12\].
Therefore the cube root of 1728 is 12.

Note:
Students should note that if any number does not have pairs of 3, so these numbers will remain inside the sign of cube root only. Students can check their answer by taking cube of 12 and checking if it is equal to 1728. $ \sqrt[3]{x} $ can also be written as $ {{\left( x \right)}^{\dfrac{1}{3}}} $ . Note that $ \sqrt[3]{{{x}^{3}}}=x $ because $ \sqrt[3]{{{x}^{3}}}={{\left( {{x}^{3}} \right)}^{\dfrac{1}{3}}}={{x}^{\dfrac{3}{3}}}={{x}^{1}}=x $ .