Question

Solve the following expression : \[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\].
(a) None of these
(b) Cannot be determined
(c) \[{3^{ - \dfrac{1}{3}}}\]
(d) \[{3^{\dfrac{1}{3}}}\]

Answer 1

Hint: The given problem revolves around the concepts of algebraic solution, which entirely depends on applying the certain rules of indices by converting all the respective terms of the given expression in one format that is least divisible multiple factor for all the terms particularly; say, here we have to convert all the desire factors to its simplest form i.e. \[3\]. As a result, applying the respective rules of indices such as multiplication, dividation, etc., the desired solution is obtained.

Complete step-by-step solution:
Since, we have given the expression that,
\[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\]
Let, ‘\[x\]’ satisfies the given expression
Hence, the expression becomes
\[x = \dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\]
Since, making the numerator as well as denominator – in the multiples of ‘\[3\]’ so as to make the respective equation efficient to solve, we get
 As a result, we know that, \[{3^2} = 9\], the equation becomes
\[x = \dfrac{{{{\left[ {{{\left( 3 \right)}^2}} \right]}^{\dfrac{1}{3}}} \times {{27}^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\]
Similarly, substituting cube root of \[27 = {\left( 3 \right)^3}\], we get
\[x = \dfrac{{{{\left[ {{{\left( 3 \right)}^2}} \right]}^{\dfrac{1}{3}}} \times {{\left[ {{{\left( 3 \right)}^3}} \right]}^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\]
Hence, now by using the rules for indices that is , \[{\left( {{a^x}} \right)^y} = {a^{xy}}\] respectively, we get
\[x = \dfrac{{{{\left( 3 \right)}^{\dfrac{2}{3}}} \times {{\left[ {{{\left( 3 \right)}^3}} \right]}^{ - \dfrac{1}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\]
Similarly, for the next term (in the numerator), we get
\[x = \dfrac{{{{\left( 3 \right)}^{\dfrac{2}{3}}} \times {{\left( 3 \right)}^{ - \dfrac{3}{2}}}}}{{{3^{\dfrac{1}{6}}} \times {3^{ - \dfrac{2}{3}}}}}\]
Hence, now, since taking the all of the denominators (present), in the numerator, where the indices of the certain terms becomes inverse of its respective sign such as positive becomes negative and negative becomes positive respectively (according to the mathematics) [also, this is one of the rules of indices for dividation i.e. \[\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}\]], we get
\[x = \dfrac{{{{\left( 3 \right)}^{\dfrac{2}{3}}} \times {{\left( 3 \right)}^{ -\dfrac{3}{2}}} \times {3^{ - \dfrac{1}{6}}}}}{{{3^{ - \dfrac{2}{3}}}}}\]
Similarly, taking the last term of denominator in the numerator, we get
\[x = {\left( 3 \right)^{\dfrac{2}{3}}} \times {\left( 3 \right)^{ - \dfrac{3}{2}}} \times {3^{ - \dfrac{1}{6}}} \times {3^{\dfrac{2}{3}}}\]
Now,
Hence, again using the certain rules for indices such as, \[{\left( a \right)^x} \times {\left( a \right)^y} = {\left( a \right)^{x + y}}\] that is, when the base of certain indices are same with different respective power, then there will be the addition in its respective power;
Hence, the equation becomes
 \[x = {\left( 3 \right)^{\dfrac{2}{3} + \left( { - \dfrac{3}{2}} \right)}} \times {3^{ - \dfrac{1}{6} + \dfrac{2}{3}}}\]
Solving the equation mathematically, we get
\[ x = {\left( 3 \right)^{\dfrac{2}{3} - \dfrac{3}{2}}} \times {3^{\dfrac{{ - 1 + 4}}{6}}}\]
\[\Rightarrow x = {\left( 3 \right)^{\dfrac{{4 - 9}}{6}}} \times {3^{\dfrac{3}{6}}} = {\left( 3 \right)^{ - \dfrac{5}{6}}} \times {3^{\dfrac{3}{6}}}\]
Hence, by using rules of indices (as discussed above), we get
\[\Rightarrow x = {\left( 3 \right)^{\dfrac{{ - 5 + 3}}{6}}}\]
\[\Rightarrow x = {\left( 3 \right)^{\dfrac{{ - 2}}{6}}}\]
Therefore, the solution of the given expression inhibits that,
\[\Rightarrow x = {\left( 3 \right)^{ - \dfrac{1}{3}}}\]
\[\therefore \] The option (c) is absolutely correct.

Note: One must able to know all the rules of indices when the certain terms are (especially) in multiplication, dividation stage such as \[{a^m} \times {a^n} = {a^{m + n}}\], \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\], \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], etc. (also, discussed briefly) which seems the most important thing in solving any algebraic solutions to get the desired output. Also, remember \[{a^m} = \dfrac{1}{{{a^{ - m}}}}\], \[{anything}^{0} = 1\], \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\], so as to be sure of our final answer.