Question

please answer$[(\frac{2}{3})^{2}]^{3}multiply(\frac{1}{3})^{-4}multiply 3^{-4} multiply 6^{-1}$

Answer 1

We have been asked to calculate the following expression:

$\longrightarrow \Bigg[\bigg(\dfrac{2}{3}\bigg)^{2}\Bigg]^{3} \times \bigg(\dfrac{1}{3}\bigg)^{-4} \times 3^{-4} \times 6^{-1}$

Hint: Try to simplify the expression.

• Expression - A mathematical symbol, or combination of symbols, represent a value, or relation. Example: $2 + 2 = 4$.

Calculation:

Let's start solving our problem and understanding every step to achieve our end result.

$\Bigg[\bigg(\dfrac{2}{3}\bigg)^{2}\Bigg]^{3} \times \bigg(\dfrac{1}{3}\bigg)^{-4} \times 3^{-4} \times 6^{-1}$

Step 1. Calculating the power of the power.

$\implies \bigg(\dfrac{2}{3}\bigg)^{6} \times \bigg(\dfrac{1}{3}\bigg)^{-4} \times 3^{-4} \times 6^{-1}$

Step 2. When raising a fraction to the power, raise the numerator and denominator each to the power.

$\implies \dfrac{2^{6}}{3^{6}} \times \bigg(\dfrac{1}{3}\bigg)^{-4} \times 3^{-4} \times 6^{-1}$

Step 3. Calculate the power.

$\implies \dfrac{64}{729} \times \bigg(\dfrac{1}{3}\bigg)^{-4} \times 3^{-4} \times 6^{-1}$

Step 4. When raising a fraction to the power, raise the numerator and denominator each to the power.

$\implies \dfrac{64}{729} \times \dfrac{1^{-4}}{3^{-4}} \times 3^{-4} \times 6^{-1}$

Step 5. If the exponent is negative, change it to a fraction.

$\implies \dfrac{64}{729} \times \dfrac{\frac{1}{1^{4}}}{3^{-4}} \times 3^{-4} \times 6^{-1}$

Step 6. Arrange the expression of the complex fraction.

$\implies \dfrac{64}{729} \times \dfrac{1}{1^{4} \times 3^{-4}} \times 3^{-4} \times 6^{-1}$

Step 7. Arrange the terms multiplied by fractions.

$\implies \dfrac{64 \times 3^{-4} \times 6^{1}}{729(1^{4} \times 3^{-4})}$

Step 8. Calculate the fraction.

$\implies \dfrac{64}{2 \times 3^{7}}$

Step 9. Reduce the fraction.

$\implies \dfrac{32}{3^{7}}$

Step 10. Calculate the power.

$\implies \dfrac{32}{2187}$

Final answer:

$\boxed{\Bigg[\bigg(\dfrac{2}{3}\bigg)^{2}\Bigg]^{3} \times \bigg(\dfrac{1}{3}\bigg)^{-4} \times 3^{-4} \times 6^{-1} = \dfrac{32}{2187}}$

Answer 2

Given: The equation is

$\left[\left(\frac{2}{3}\right)^{2}\right]^{3} \text { multiply }\left(\frac{1}{3}\right)^{-4} \text { multiply } 3^{-4} \text { multiply } 6^{-1}$

We have to solve the above equation.

• By using the Bodmas rule, we are solving the above equation.
• As we know that the Bodmas rule is used to remember the order of operations to be followed while solving expressions in mathematics.

Where,

$\begin{array}{l}\mathrm{B}=\text{brackets}\\\mathrm{O}=\text { order of powers or rules } \\\mathrm{D}=\text { division } \\\mathrm{M}=\text { multiplication } \\\mathrm{A}=\text { addition } \\\mathrm{S}=\text { subtraction }\end{array}$

We are solving in the following way:

We have,

$\left[\left(\frac{2}{3}\right)^{2}\right]^{3} \text { multiply }\left(\frac{1}{3}\right)^{-4} \text { multiply } 3^{-4} \text { multiply } 6^{-1}$

$=>[\frac{4}{9} ]^{3} \times 81 \times\frac{1}{81} \times\frac{1}{6} \\\\=>\frac{64}{729} \times 81 \times\frac{1}{81} \times\frac{1}{6}\\\\=>\frac{64}{9} \times\frac{1}{81} \times\frac{1}{6}\\\\=>\frac{32}{2187}$

Hence, the solution of the above equation is$\frac{32}{2187}$.