52728
= 3 x 2^3 x 13^3, and (2^3 x 13^3) is a perfect cube
so dividing 52728 by 3 would leave a quotient of 17576 which is 2^3 x 13^3 or 26^3
thus the smallest such number as a divisor is 3
and the cube root of the quotient is 26
= 3 x 2^3 x 13^3, and (2^3 x 13^3) is a perfect cube
so dividing 52728 by 3 would leave a quotient of 17576 which is 2^3 x 13^3 or 26^3
thus the smallest such number as a divisor is 3
and the cube root of the quotient is 26
Divide 52728 by the smallest number so that the quotient is a perfect cube
Step 1: Prime factorize 52728:
52728 = 2 × 2 × 2 × 3 × 3 × 463
Step 2: Group the prime factors into triples to find the perfect cubes:
2 × 2 × 2 = 8 (2^3)
3 × 3 × 463 = 4179 (3^3 × 463)
Step 3: Calculate the smallest number by multiplying the prime factors that make up the perfect cubes:
Smallest number = 2 × 2 × 2 × 3 × 3 × 463 = 52728
Therefore, the smallest number by which 52728 can be divided so that the quotient is a perfect cube is 52728 itself.
Now, let's find the cube root of the quotient:
Quotient = 52728 / 52728 = 1
The cube root of 1 is also 1.
So, the quotient is 1 and its cube root is 1.