Question

Calculate the value of $\sqrt[3]{{\dfrac{{ - 192}}{{81}}}}$.
(a) $- \dfrac{5}{3}$
(b) $\dfrac{{ - 4}}{3}$
(c) $\dfrac{3}{2}$
(d) $\dfrac{{13}}{9}$

Answer 1

Here, we need to simplify the given expression. We will rewrite the expression inside the cube root as a power of 3 using divisibility tests and rules of exponents. Then, we will use rules of exponents to simplify the given expression. Cube root is the inverse of the calculation of the cube of a number.

Formula Used:
We will use the following formulas:
1.If two or more numbers with same base and different exponents are multiplied, the product can be written as ${a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}$.
2.If two or more numbers with different bases and same exponent are multiplied, the product can be written as ${a^m} \times {b^m} = {\left( {ab} \right)^m}$.
3.If two or more numbers with different bases and same exponent are divided, the quotient can be written as $\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}$.
4.If a number with an exponent is raised to another exponent, then the exponents are multiplied. This can be written as ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$.

Complete step-by-step answer:
We will rewrite the expression inside the cube root and use rules of exponents to simplify the expression.
192 is the product of 64 and 3.
81 is the product of 27 and 3.
Simplifying the given expression, we get
$\sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{ - 64}}{{27}}}}$
First, we will express the numbers 64 and 27 as a power of 3.
Let us use the divisibility tests of 2, 3, 5, etc. to find the factors of 64.
First, we will check the divisibility by 2.
We know that a number is divisible by 2 if it is an even 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 64 has 4 at the unit’s place.
Therefore, the number 64 is divisible by 2.
Dividing 64 by 2, we get
$\dfrac{{64}}{2} = 32$
Now, the number 32 has 2 at the unit’s place.
Therefore, the number 32 is divisible by 2.
Dividing 32 by 2, we get
$\dfrac{{32}}{2} = 16$
We know that 16 is the product of 4 and 4.
Therefore, we can rewrite the number 64 as
$64 = 2 \times 2 \times 4 \times 4$
Multiplying 2 by 2 in the expression, we get
$64 = 4 \times 4 \times 4$
Applying the rule of exponent ${a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}$, we can rewrite 64 as
$\begin{array}{l} \Rightarrow 64 = {4^{1 + 1 + 1}}\\\ \Rightarrow 64 = {4^3}\end{array}$
Therefore, we have expressed 64 as 4 raised to the power 3.
Now, we will use the divisibility tests of 2, 3, 5, etc. to find the factors of 27.
We can observe that the number 27 has 7 at the unit’s place.
Therefore, the number 27 is not divisible by 2.
Next, we will check the divisibility by 3.
A number is divisible by 3 if the sum of its digits is divisible by 3.
We will add the digits of the number 27.
Thus, we get
$2 + 7 = 9$
Since the number 9 is divisible by 3, the number 27 is divisible by 3.
Dividing 27 by 3, we get
$\dfrac{{27}}{3} = 9$
We know that 9 is the product of 3 and 3.
Therefore, we can rewrite the number 27 as
$27 = 3 \times 3 \times 3$
Applying the rule of exponent ${a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}$, we can rewrite 27 as
$\begin{array}{l} \Rightarrow 27 = {3^{1 + 1 + 1}}\\\ \Rightarrow 27 = {3^3}\end{array}$
Therefore, we have expressed 27 as 3 raised to the power 3.
Finally, we will simplify the given expression.
Substituting $64 = {4^3}$ and $27 = {3^3}$ in the equation $\sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{ - 64}}{{27}}}}$, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{ - {4^3}}}{{{3^3}}}}}$
Rewriting the expression, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{ - 1 \times {4^3}}}{{{3^3}}}}}$
We know that the product $- 1 \times - 1 \times - 1$ is equal to $- 1$.
Therefore, we can rewrite the expression as
$\begin{array}{l} \Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{\left( { - 1 \times - 1 \times - 1} \right) \times {4^3}}}{{{3^3}}}}}\\\ \Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{{{\left( { - 1} \right)}^3} \times {4^3}}}{{{3^3}}}}}\end{array}$
Applying the rule of exponent ${a^m} \times {b^m} = {\left( {ab} \right)^m}$, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{{{\left( { - 1 \times 4} \right)}^3}}}{{{3^3}}}}}$
Multiplying the terms, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{\dfrac{{{{\left( { - 4} \right)}^3}}}{{{3^3}}}}}$
Applying the rule of exponent $\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}$, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \sqrt[3]{{{{\left( {\dfrac{{ - 4}}{3}} \right)}^3}}}$
Rewriting the equation, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = {\left[ {{{\left( {\dfrac{{ - 4}}{3}} \right)}^3}} \right]^{\dfrac{1}{3}}}$
Rule of exponent: If a number with an exponent is raised to another exponent, then the exponents are multiplied. This can be written as ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$.
Applying the rule of exponent ${\left( {{a^m}} \right)^n} = {a^{m \times n}}$, we get
$\Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = {\left( {\dfrac{{ - 4}}{3}} \right)^{3 \times \dfrac{1}{3}}}$
Therefore, we get
$\begin{array}{l} \Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = {\left( {\dfrac{{ - 4}}{3}} \right)^1}\\\ \Rightarrow \sqrt[3]{{\dfrac{{ - 192}}{{81}}}} = \dfrac{{ - 4}}{3}\end{array}$
Therefore, we get the value of the expression $\sqrt[3]{{\dfrac{{ - 192}}{{81}}}}$ as $\dfrac{{ - 4}}{3}$.
Thus, the correct option is option (b).

Note: We expressed 64 and 27 as 4 raised to the power 3, and 3 raised to the power 3, respectively. Here, 64 is called the cube of 4, and 27 is called the cube of 3. When a number is raised to the power 3, the resulting product is its cube.
We simplified ${\left( {\dfrac{{ - 4}}{3}} \right)^1}$ as $\dfrac{{ - 4}}{3}$. This is because any number raised to the exponent 1 is equal to itself.