Question

Find the value of cube root \[\sqrt[3]{\dfrac{-64}{1331}}\]
\[\begin{align}
  & (A)\text{ }\dfrac{-4}{11} \\
 & \text{(B) }\dfrac{4}{11} \\
 & \text{(C) }\dfrac{-14}{11} \\
 & \text{(D) }\dfrac{-4}{1} \\
\end{align}\]

Answer 1

Hint:First of all, we have to compare \[\sqrt[3]{\dfrac{-64}{1331}}\] with \[\sqrt[n]{\dfrac{a}{b}}\]. Now we will convert \[\sqrt[n]{\dfrac{a}{b}}\] in to \[{{\left( \dfrac{a}{b} \right)}^{\dfrac{1}{n}}}\]. Now by using the formula \[{{\left( \dfrac{a}{b} \right)}^{n}}=\]\[\dfrac{{{a}^{n}}}{{{b}^{n}}}\], we have to proceed. Now by using the formula that \[{{({{a}^{m}})}^{n}}={{a}^{mn}}\]we will proceed further. Now we will get the required result.

Complete step-by-step answer:
Before solving the question, we should know that \[\sqrt[n]{\dfrac{a}{b}}\] can be written as \[{{\left( \dfrac{a}{b} \right)}^{\dfrac{1}{n}}}\]. In the question, we were given to find the value of \[\sqrt[3]{\dfrac{-64}{1331}}\]. So, by comparing \[\sqrt[3]{\dfrac{-64}{1331}}\] with \[\sqrt[n]{\dfrac{a}{b}}\]we can write \[\sqrt[3]{\dfrac{-64}{1331}}\]as \[{{\left( \dfrac{-64}{1331} \right)}^{\dfrac{1}{3}}}\].
\[\Rightarrow \sqrt[3]{\dfrac{-64}{1331}}={{\left( \dfrac{-64}{1331} \right)}^{\dfrac{1}{3}}}....(1)\].
We know that \[{{\left( \dfrac{a}{b} \right)}^{n}}\] can be written as \[\dfrac{{{a}^{n}}}{{{b}^{n}}}\].
So, we rewrite equation (1).
\[\Rightarrow {{\left( \dfrac{-64}{1331} \right)}^{\dfrac{1}{3}}}=\dfrac{{{(-64)}^{\dfrac{1}{3}}}}{{{(1331)}^{\dfrac{1}{3}}}}....(2)\]
We know that 64 can be written as a product of 4, 4 and 4.
\[\Rightarrow 64=\text{4}\times \text{4}\times \text{4=}{{\text{4}}^{3}}\]
In the same way, -6 can be written as a product of -4, -4 and -4.
\[\Rightarrow -64=(-\text{4)}\times (-\text{4)}\times (-\text{4)=(-4}{{\text{)}}^{3}}.....\text{(3)}\]
We know that 1331 can be written as a product of 11, 11 and 11.
\[\Rightarrow 1331=11\times \text{11}\times \text{11=(11}{{\text{)}}^{3}}.....\text{(4)}\]
Now we will substitute equation (3) and equation (4) ion equation (2).
\[\Rightarrow \dfrac{{{(-64)}^{\dfrac{1}{3}}}}{{{(1331)}^{\dfrac{1}{3}}}}=\dfrac{{{(-{{4}^{3}})}^{\dfrac{1}{3}}}}{{{({{11}^{3}})}^{1/3}}}....(5)\]
We know that \[{{({{a}^{m}})}^{n}}\] is equal to \[{{a}^{mn}}\].
So, we get
\[\Rightarrow {{(-{{4}^{3}})}^{\dfrac{1}{3}}}={{(-4)}^{3\times \left( \dfrac{1}{3} \right)}}={{(-4)}^{1}}=-4...(6)\]
In the similar manner,
\[\Rightarrow {{({{11}^{3}})}^{1/3}}={{11}^{3\times \left( \dfrac{1}{3} \right)}}={{11}^{1}}=11...(7)\].
Now we will substitute equation (6) and equation (7) in equation (5).
 \[\Rightarrow \dfrac{{{(-{{4}^{3}})}^{\dfrac{1}{3}}}}{{{({{11}^{3}})}^{\dfrac{1}{3}}}}=\dfrac{-4}{11}\].
Hence, we get the value of \[\sqrt[3]{\dfrac{-64}{1331}}\] is equal to \[\dfrac{-4}{11}\].
So, option (A) is correct.

Note: Students must be careful while using exponential formulae. Students may be confused that the square root of a negative integer will not give a real number. In the same way, students may think the cube root of a negative integer also will not give a real number. This is a misconception. We should remember that the square root of a negative integer will not give a real number but the cube root of a negative integer will result in an integer. .