Question

How do you evaluate \[{{\left( -\dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\]?

Answer 1

Hint: Assume the given expression as ‘E’. Write the base of ‘E’ as \[{{\left( -\dfrac{27}{8} \right)}^{\dfrac{4}{3}}}={{\left( -1\times \dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\]. Now, write the numbers in the numerator and the denominator into their exponential forms by using the prime factorization. In the next step, apply the formula of exponents and powers as: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] to simplify the value of the expression. Write \[-1={{\left( -1 \right)}^{3}}\] and simplify further to get the answer.

Complete step-by-step answer:
Here, we have been provided with the expression \[{{\left( -\dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\] and we are asked to evaluate it.
Now, let us assume the value of the given expression as E. So, we have,
\[\Rightarrow E={{\left( -\dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\]
We can write the base, i.e. \[{{\left( -\dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\] as \[{{\left( -1\times \dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\], so we get,
\[\Rightarrow E={{\left( -1\times \dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\]
Breaking the terms by using the formula: - \[{{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}\], we get,
\[\Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times {{\left( \dfrac{27}{8} \right)}^{\dfrac{4}{3}}}\]
Now, we need to convert the fraction $\dfrac{27}{8}$ into the exponential form by using the prime factorization method, so writing both the numerator and the denominator as the product of their primes we get the expression as:
\[\Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times {{\left( \dfrac{3\times 3\times 3}{2\times 2\times 2} \right)}^{\dfrac{4}{3}}}\]
Converting the above expression into the exponential form we get,
\[\Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times {{\left( {{\left( \dfrac{3}{2} \right)}^{3}} \right)}^{\dfrac{4}{3}}}\]
Using the formula of exponents given as: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
  & \Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times {{\left( \dfrac{3}{2} \right)}^{3\times \dfrac{4}{3}}} \\
 & \Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times {{\left( \dfrac{3}{2} \right)}^{4}} \\
 & \Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times \left( \dfrac{3\times 3\times 3\times 3}{2\times 2\times 2\times 2} \right) \\
 & \Rightarrow E={{\left( -1 \right)}^{\dfrac{4}{3}}}\times \dfrac{81}{16} \\
\end{align}\]
Now, we can write $-1={{\left( -1 \right)}^{3}}$, so the expression becomes,
\[\begin{align}
  & \Rightarrow E={{\left( {{\left( -1 \right)}^{3}} \right)}^{\dfrac{4}{3}}}\times \dfrac{81}{16} \\
 & \Rightarrow E={{\left( -1 \right)}^{3\times \dfrac{4}{3}}}\times \dfrac{81}{16} \\
 & \Rightarrow E={{\left( -1 \right)}^{4}}\times \dfrac{81}{16} \\
\end{align}\]
Here, -1 will be multiplied 4 times and we know that $\left( -1 \right)\times \left( -1 \right)=1$, so we get,
\[\begin{align}
  & \Rightarrow E=1\times \dfrac{81}{16} \\
 & \Rightarrow E=\dfrac{81}{16} \\
\end{align}\]
Hence, the above expression represents the simplified form of the given exponential expression.

Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}},{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}},{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] and \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] because they are used everywhere. Remember that: - \[{{\left( -1 \right)}^{n}}=1\] when ‘n’ is an even integer and \[{{\left( -1 \right)}^{n}}=-1\] when ‘n’ is an odd integer. You must know how to write a number as the product of its primes.