Evaluate (1-3((3^4)/4)^-1(3^3)/4)^10

${(1 - 3 \cdot {(\frac{3^{4}}{4})}^{- 1} \cdot \frac{3^{3}}{4})}^{10}$

Step 1

Simplify each term.

Raise $3$ to the power of $3$ .

${(1 - 3 \cdot {(\frac{3^{4}}{4})}^{- 1} \cdot \frac{27}{4})}^{10}$

Raise $3$ to the power of $4$ .

${(1 - 3 \cdot {(\frac{81}{4})}^{- 1} \cdot \frac{27}{4})}^{10}$

Change the sign of the exponent by rewriting the base as its reciprocal.

${(1 - 3 \cdot \frac{4}{81} \cdot \frac{27}{4})}^{10}$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

${(1 + 3 (- 1) \cdot \frac{4}{81} \cdot \frac{27}{4})}^{10}$

Factor $3$ out of $81$ .

${(1 + 3 \cdot - 1 \cdot \frac{4}{3 \cdot 27} \cdot \frac{27}{4})}^{10}$

Cancel the common factor.

${(1 + 3 \cdot - 1 \cdot \frac{4}{3 \cdot 27} \cdot \frac{27}{4})}^{10}$

Rewrite the expression.

${(1 - 1 \cdot \frac{4}{27} \cdot \frac{27}{4})}^{10}$

Rewrite $- 1 (\frac{4}{27})$ as $- (\frac{4}{27})$ .

${(1 - (\frac{4}{27}) \cdot \frac{27}{4})}^{10}$

Cancel the common factor of $4$ .

Move the leading negative in $- (\frac{4}{27})$ into the numerator.

${(1 + \frac{- 4}{27} \cdot \frac{27}{4})}^{10}$

Factor $4$ out of $- 4$ .

${(1 + \frac{4 (- 1)}{27} \cdot \frac{27}{4})}^{10}$

Cancel the common factor.

${(1 + \frac{4 \cdot - 1}{27} \cdot \frac{27}{4})}^{10}$

Rewrite the expression.

${(1 + \frac{- 1}{27} \cdot 27)}^{10}$

Cancel the common factor of $27$ .

Cancel the common factor.

${(1 + \frac{- 1}{27} \cdot 27)}^{10}$

Rewrite the expression.

${(1 - 1)}^{10}$

Step 2

Simplify the expression.

Subtract $1$ from $1$ .

$0^{10}$

Raising $0$ to any positive power yields $0$ .

$0$

Evaluate (1-3*((3^4)/4)^-1*(3^3)/4)^10