Question

How do you evaluate \[ - {81^{\dfrac{3}{4}}}\] ?

Answer 1

Hint:We can see this problem is from indices and powers. This number given is having $81$ as base and $\left( {\dfrac{3}{4}} \right)$ as power. But since we have to simplify this we can write $\left( {\dfrac{3}{4}} \right)$ as the cube of the fourth-root of the given number in base. So, we first have to find out the square of the base number, $4$ and then the cube root of the square of $4$. The result thus obtained will be the answer to the given question.

Complete step by step answer:
The given question requires us to simplify \[ - {81^{\dfrac{3}{4}}}\]. This is of the form ${a^{\left( {\dfrac{m}{n}} \right)}}$. But we can rewrite the expression using the laws of indices and powers ${a^{\left( {\dfrac{m}{n}} \right)}} = {\left( {{a^{\dfrac{1}{n}}}} \right)^m}$.Thus we will apply same on the question above, we get,
$ - {81^{\dfrac{3}{4}}} = - {\left( {{{81}^{\dfrac{1}{4}}}} \right)^3}$
Now we first solve the power inside the bracket.
$ \Rightarrow - {\left( {{{81}^{\dfrac{1}{4}}}} \right)^3}$

Now we should know that the power $\left( {\dfrac{1}{4}} \right)$ represents the fourth-root of the entity. To find the fourth root of $\left( {81} \right)$, we have to first factorise the number.
$81 = 3 \times 3 \times 3 \times 3 = \left( {{3^4}} \right)$
So, we get, ${\left( {81} \right)^{\dfrac{1}{4}}} = {\left( {{3^4}} \right)^{\dfrac{1}{4}}} = 3$
Hence,
$ \Rightarrow - {\left( {{{81}^{\dfrac{1}{4}}}} \right)^3} = - {3^3}$
Now, we will have to compute the cube of the number $3$.
We know that ${a^3} = a \times a \times a$. So, we get, ${3^3} = 3 \times 3 \times 3 = 27$.
$ \therefore - {\left( {{{81}^{\dfrac{1}{4}}}} \right)^3} = - 27$

Hence, we get the value of \[ - {81^{\dfrac{3}{4}}}\] as $\left( { - 27} \right)$.

Note:These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be moulded according to our convenience while solving the problem. Also note that cube-root, square-root are fractions with 1 as numerator and respective root in denominator.