Question

By what number should we multiply $ - {6^{ - 3}} $ so that the product is $ {\left( {\dfrac{1}{3}} \right)^5} $ .

Answer 1

Hint: We have to find the number which we multiply with $ {( - 6)^{ - 3}} $ and get the product $ {\left( {\dfrac{1}{3}} \right)^5} $ . We will assume that number to be ‘x’. Then, we will multiply $ {\left( { - 6} \right)^{ - 3}} $ with $ x $ in L.H.S. and write $ {\left( {\dfrac{1}{3}} \right)^5} $ in R.H.S. One term has the power of negative in the question. So, we can write it as its reciprocal with the positive power like $ \left( {{a^{ - n}} = \dfrac{1}{{{a^n}}}} \right) $ .

Complete step-by-step answer:
Assume the number which we multiply with $ {\left( { - 6} \right)^{ - 3}} $ and product will come $ {\left( {\dfrac{1}{3}} \right)^5} $ as ‘x’.
Now, in L.H.S. we will multiply $ {\left( { - 6} \right)^{ - 3}} $ with $ x $ and in R.H.S. We will write $ {\left( {\dfrac{1}{3}} \right)^5} $ .
\[{\left( { - 6} \right)^{ - 3}} \times x = {\left( {\dfrac{1}{3}} \right)^5}\]
If the power of a digit is negative we can write it as its reciprocal but with positive power. So, we will rewrite $ {\left( { - 6} \right)^{ - 3}} $ as $ {\left( {\dfrac{1}{{ - 6}}} \right)^3} $ in the above equation.
 $ {\left( {\dfrac{1}{{ - 6}}} \right)^3} \times x = {\left( {\dfrac{1}{3}} \right)^5} $
 $ x = \dfrac{1}{{{3^5}}} \times {( - 6)^3} $
 $ x = \dfrac{{ - {6^3}}}{{{3^5}}} $
We can write 6 as $ 2 \times 3 $ .
 $ x = - \dfrac{{{{\left( {2 \times 3} \right)}^3}}}{{{3^5}}} $
By using the property $ {\left( {a \times b} \right)^n} = {a^n} \times {b^n} $ , we will rewrite the above equation.
 $ x = - \dfrac{{{2^3} \times {3^3}}}{{{3^5}}} $
 $ x = - \dfrac{{{2^3}}}{{{3^2}}} $
 $ x = - \dfrac{8}{9} $
So, we should divide $ - \dfrac{8}{9} $ with $ {\left( { - 6} \right)^{ - 3}} $ so the product comes as $ {\left( {\dfrac{1}{3}} \right)^5} $ .
So, the correct answer is “$ - \dfrac{8}{9} $”.

Note: We note that $ {a^{ - n}} $ is the reciprocal of $ {a^n} $ and vice-versa. We have used this property in the solution. L.H.S. is informal shorthand for the left-hand side of an equation. Similarly, R.H.S. is the right-hand side of the equation. The properties related to algebraic expression must be learned to solve these types of questions. In this case, we have used some of them. We should be careful that base and exponent should never be zero simultaneously. We should always remember divisibility rules while solving the question.