The cube root of −729 is:

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−19

−9

−39

−3

answer is B.

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Detailed Solution

Concept- As a product of a negative and a positive integer, let's rewrite the provided number.
When we rewrite the number -729, we obtain -729=-1×729
We are aware that the square of one is one.
As a result, the equation 729=-1×729 can be written as -729=-(1)3×729
We'll now write -729 as a power of 3.
Let's find the factors of -729 using the tests for 2, 3, 5, etc. divisibility.
We shall first examine the divisibility by two.
We are aware that an even number, not an odd one, is one that can be divided by two.
This indicates that any integer is divisible by 2 if the unit is replaced by one of the digits 2, 4, 6, 8, or 0.
We shall then examine the divisibility by three.
A number is only three digits divisible if the sum of its digits is three digits divisible.
The digits of the number 729 will be added.
So, we obtain 7+2+9=18.
The number 729 is divisible by three because the number 18 is also divisible by three.
729 divided by 3 equals 7293, or 243.
We'll then add the numerals to make 243.
The result is 2+4+3=9.
Since the number 9 is divisible by 3,
So, too is the number 243.
We get

\frac{243}{3}

=81 when we divide 243 by 3.
We'll then add the digits to make 81.
So, we obtain 8+1=9.
Since the number 9 is divisible by 3, so too is the number 81.
We get 813=27 when we divide 81 by 3.
We are aware that the sum of 3 and 9 equals 27.
As a result, 729 can now be written as 729=3×3×3×3×9.
When we multiply each pair of threes by itself, we obtain 729=9×9×9
When we multiply each pair of threes by itself, we obtain 729=9×9×9
We are aware that the product of two or more numbers with the same base but different exponents can be expressed as abacad=ab+c+d. This is known as the exponent rule.
Thus, 729 can be rewritten as 729=