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Hint: In the above given question, we are asked to evaluate the expression ${(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}}$. While evaluation, do not try to jump over the steps, as if any step is missed the solution so obtained will not be appropriate. Also remember the rule that in multiplication if the base is the same, the powers are added.

We have the given expression as

${(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}}$

Now we know that, we can write

${(27{x^3})^{\dfrac{1}{3}}}$as${({(3x)^3})^{\dfrac{1}{3}}}$ … (1)

And ${(81{x^3})^{\dfrac{1}{3}}}$as${({(3)^4})^{\dfrac{1}{3}}} \times {({(y)^3})^{\dfrac{1}{3}}}$ … (2)

Therefore, from these we can rewrite the given expression as

$ \Rightarrow {(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}} = {({(3x)^3})^{\dfrac{1}{3}}} \times {({(3)^4})^{\dfrac{1}{3}}} \times {({(y)^3})^{\dfrac{1}{3}}}$

$ = {(3x)^{\dfrac{3}{3}}} \times {(3)^{\dfrac{4}{3}}} \times {(y)^{\dfrac{3}{3}}}$

$ = {(3x)^1} \times {(3)^{\dfrac{4}{3}}} \times {(y)^1}$

After solving the expression, we get

$ = {(3x)^1} \times {(3)^1}{(3)^{\dfrac{1}{3}}} \times {(y)^1}$

$ = {(3x)^1} \times {(3)^1}{(3)^{\dfrac{1}{3}}} \times {(y)^1}$

$ = 3x \times 3 \times {(3)^{\dfrac{1}{3}}} \times y$

$ = x \times {3^2} \times {3^{\dfrac{1}{3}}} \times y$

$ = x \times {3^{\dfrac{{6 + 1}}{3}}} \times y$

$ = x \times {3^{\dfrac{7}{3}}} \times y$

$ = {3^{\dfrac{7}{3}}}xy $

Hence, the required solution is:

${(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}} = {3^{\dfrac{7}{3}}}xy$

Note: Whenever we face such types of problems, break the given terms in the simplest form possible and then evaluate the given expression to obtain an optimum solution.

We have the given expression as

${(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}}$

Now we know that, we can write

${(27{x^3})^{\dfrac{1}{3}}}$as${({(3x)^3})^{\dfrac{1}{3}}}$ … (1)

And ${(81{x^3})^{\dfrac{1}{3}}}$as${({(3)^4})^{\dfrac{1}{3}}} \times {({(y)^3})^{\dfrac{1}{3}}}$ … (2)

Therefore, from these we can rewrite the given expression as

$ \Rightarrow {(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}} = {({(3x)^3})^{\dfrac{1}{3}}} \times {({(3)^4})^{\dfrac{1}{3}}} \times {({(y)^3})^{\dfrac{1}{3}}}$

$ = {(3x)^{\dfrac{3}{3}}} \times {(3)^{\dfrac{4}{3}}} \times {(y)^{\dfrac{3}{3}}}$

$ = {(3x)^1} \times {(3)^{\dfrac{4}{3}}} \times {(y)^1}$

After solving the expression, we get

$ = {(3x)^1} \times {(3)^1}{(3)^{\dfrac{1}{3}}} \times {(y)^1}$

$ = {(3x)^1} \times {(3)^1}{(3)^{\dfrac{1}{3}}} \times {(y)^1}$

$ = 3x \times 3 \times {(3)^{\dfrac{1}{3}}} \times y$

$ = x \times {3^2} \times {3^{\dfrac{1}{3}}} \times y$

$ = x \times {3^{\dfrac{{6 + 1}}{3}}} \times y$

$ = x \times {3^{\dfrac{7}{3}}} \times y$

$ = {3^{\dfrac{7}{3}}}xy $

Hence, the required solution is:

${(27{x^3})^{\dfrac{1}{3}}} \times {(81{y^3})^{\dfrac{1}{3}}} = {3^{\dfrac{7}{3}}}xy$

Note: Whenever we face such types of problems, break the given terms in the simplest form possible and then evaluate the given expression to obtain an optimum solution.

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