Question

Find the cube root of 1728 using prime factorization.

Answer 1

Hint: To solve this particular type of question we need to find the prime factors of 1728 and put them into groups of three to find the cube root of each of the groups and then multiply them to get the desired answer.

Complete step-by-step answer:
Prime factors of 1728 are $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3

\times 3$

Therefore cube root of 1728 $ \to $

 $

  {1728^{\dfrac{1}{3}}} = {\left( {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times

3 \times 3} \right)^{\dfrac{1}{3}}} \\

   = {\left( {2 \times 2 \times 2} \right)^{\dfrac{1}{3}}} \times {\left( {2 \times 2 \times 2}

\right)^{\dfrac{1}{3}}} \times {\left( {3 \times 3 \times 3} \right)^{\dfrac{1}{3}}} \\

   = 2 \times 2 \times 3 = 12 \\

$

Note: Remember to recall the method of prime factorization in order to solve such types of questions. Note that for finding the cube root we take the prime factor that occurs three times only once and multiply. Always consider taking prime factors to reduce confusion while solving.