Question

The cube root of \[ - 729\] is
(a) \[ - 19\]
(b) \[ - 9\]
(c) \[ - 39\]
(d) \[ - 3\]

Answer 1

Hint:
Here, we need to find the cube root of \[ - 729\]. First, we will rewrite the given number as a product of a negative and positive integer. Then, we will use the factors of the two numbers to express them as a power of 3. Finally, we will take the cube root of the given number and simplify the right hand side to get the required value.

Formula Used:
We will use the rule of the exponent, \[{a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}\].

Complete step by step solution:
Let us rewrite the given number as a product of a negative and positive integer.
Rewriting the number \[ - 729\], we get
\[ - 729 = - 1 \times 729\]
We know that the cube of \[ - 1\] is \[ - 1\].
Therefore, we can rewrite the equation \[ - 729 = - 1 \times 729\] as
\[ \Rightarrow - 729 = {\left( { - 1} \right)^3} \times 729\]
Now, we will express 729 as a power of 3.
Let us use the divisibility tests of 2, 3, 5, etc. to find the factors of 729.
First, we will check the divisibility by 2.
We know that if a number is divisible by 2 then it is an even number not an odd number.
This means that any number that has one of the digits 2, 4, 6, 8, or 0 in the unit’s place, is divisible by 2.
We can observe that the number 729 has 9 at the unit’s place.
Therefore, the number 729 is not divisible by 2.
Next, we will check the divisibility by 3.
A number is divisible by 3 only if the sum of its digits is divisible by 3.
We will add the digits of the number 729.
Thus, we get
\[7 + 2 + 9 = 18\]
Since the number 18 is divisible by 3, the number 729 is divisible by 3.
Dividing 729 by 3, we get
\[\dfrac{{729}}{3} = 243\]
Next, we will add the digits of the number 243.
Thus, we get
\[2 + 4 + 3 = 9\]
Since the number 9 is divisible by 3, the number 243 is divisible by 3.
Dividing 243 by 3, we get
\[\dfrac{{243}}{3} = 81\]
Next, we will add the digits of the number 81.
Thus, we get
\[8 + 1 = 9\]
Since the number 9 is divisible by 3, the number 81 is divisible by 3.
Dividing 81 by 3, we get
\[\dfrac{{81}}{3} = 27\]
We know that 27 is the product of 3 and 9.
Therefore, we can rewrite the number 729 as
\[729 = 3 \times 3 \times 3 \times 3 \times 9\]
Multiplying the pairs of 3 with each other, we get
\[\begin{array}{l} \Rightarrow 729 = 9 \times 9 \times 9\\ \Rightarrow 729 = {9^1} \times {9^1} \times {9^1}\end{array}\]
We know that the rule of exponent says if two or more numbers with same base and different exponents are multiplied, the product can be written as \[{a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}\].
Therefore, we can rewrite 729 as
\[\begin{array}{l} \Rightarrow 729 = {9^{1 + 1 + 1}}\\ \Rightarrow 729 = {9^3}\end{array}\]
\[\therefore \] We have expressed 729 as 9 raised to the power 3.
Substituting \[729 = {9^3}\] in the equation \[ - 729 = {\left( { - 1} \right)^3} \times 729\], we get
\[ \Rightarrow - 729 = {\left( { - 1} \right)^3} \times {9^3}\]
Using the rule of exponents \[{a^m} \times {b^m} = {\left( {ab} \right)^m}\], we get
\[ \Rightarrow - 729 = {\left( { - 1 \times 9} \right)^3}\]
Multiplying the terms, we get
\[ \Rightarrow - 729 = {\left( { - 9} \right)^3}\]
Finally, we will find the cube root of \[ - 729\].
Taking the cube roots of both sides of the equation \[ - 729 = {\left( { - 9} \right)^3}\], we get
\[ \Rightarrow \sqrt[3]{{ - 729}} = \sqrt[3]{{{{\left( { - 9} \right)}^3}}}\]
Simplifying the expression, we get
\[ \Rightarrow \sqrt[3]{{ - 729}} = - 9\]
\[\therefore \] We get the cube root of the number \[ - 729\] as \[ - 9\].

Thus, the correct option is option (b).

Note:
We used the rule of exponents \[{a^m} \times {b^m} = {\left( {ab} \right)^m}\] to rewrite \[{\left( { - 1} \right)^3} \times {9^3}\] as \[{\left( { - 1 \times 9} \right)^3}\]. If two terms with different bases and same exponents are multiplied, then the result is the product of the bases, raised to the same exponent. This can be written as \[{a^m} \times {b^m} = {\left( {ab} \right)^m}\].