Cube root of 19683 by prime factorisation method
CUBE ROOT OF 19683
27 is the cube root of 19683. It means that when 27 is multiplied thrice by itself, we get the resultant number 19683.
How to find the cube root?
To find the cube root of a number, we will use the prime factorisation method. By evaluating the prime factors we will make the groups of threes of the same digits. Then we apply the cube root and choose one value from each group. After choosing the value, we will multiply those numbers. The result we get is the cube root of the required number.
How to calculate the cube root of 19683?
By using the prime factorization of 19683, we get- (3×3×3) × (3×3×3) × (3×3×3)
We can also write it as – (33×33×33) = (3×3×3)3
When we apply the cube root on (3×3×3)3, we get the number 27.
Therefore, 27 is the answer.
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Frequently Asked Questions
Q1. Is the cube root of 19683 a perfect cube?
Ans. We know that the cube root of 19683 is 27. The prime factors of 27 are (3 × 3 × 3) = (33).
When we apply the cube root on 33, we get 3 as the cube root of 27. Therefore, 27 is a perfect cube of 3.
Q2. How many real and complex cube roots does a real number have?
Ans. A real number has one real cube root and two complex cube roots.
Q3. When 19683 is divided by 27, we get the cube of which number?
Ans. When we divide 19683 by 27, we get the result 729.
When the prime factorisation is applied on 729, the prime factors we get- (3×3×3) × (3×3×3)
When we make the triplets of 3, we can write it as (33 × 33) = (3×3)3.
On applying the cube root on (3×3)3, we get 9
Hence, when 19683 is divided by 27, we get the cube of 9.