Evaluate 2/3 square root of 81- cube root of -27-4/3 cube root of 729

$\frac{2}{3} \sqrt{81} - \sqrt[3]{- 27} - \frac{4}{3} \sqrt[3]{729}$

Simplify each term.

Rewrite $81$ as $9^{2}$ .

$\frac{2}{3} \cdot \sqrt{9^{2}} - \sqrt[3]{- 27} - \frac{4}{3} \cdot \sqrt[3]{729}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{2}{3} \cdot 9 - \sqrt[3]{- 27} - \frac{4}{3} \cdot \sqrt[3]{729}$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$\frac{2}{3} \cdot (3 (3)) - \sqrt[3]{- 27} - \frac{4}{3} \cdot \sqrt[3]{729}$

Cancel the common factor.

$\frac{2}{3} \cdot (3 \cdot 3) - \sqrt[3]{- 27} - \frac{4}{3} \cdot \sqrt[3]{729}$

Rewrite the expression.

$2 \cdot 3 - \sqrt[3]{- 27} - \frac{4}{3} \cdot \sqrt[3]{729}$

Multiply $2$ by $3$ .

$6 - \sqrt[3]{- 27} - \frac{4}{3} \cdot \sqrt[3]{729}$

Rewrite $- 27$ as ${(- 3)}^{3}$ .

$6 - \sqrt[3]{{(- 3)}^{3}} - \frac{4}{3} \cdot \sqrt[3]{729}$

Pull terms out from under the radical, assuming real numbers.

$6 - - 3 - \frac{4}{3} \cdot \sqrt[3]{729}$

Multiply $- 1$ by $- 3$ .

$6 + 3 - \frac{4}{3} \cdot \sqrt[3]{729}$

Rewrite $729$ as $9^{3}$ .

$6 + 3 - \frac{4}{3} \cdot \sqrt[3]{9^{3}}$

Pull terms out from under the radical, assuming real numbers.

$6 + 3 - \frac{4}{3} \cdot 9$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{4}{3}$ into the numerator.

$6 + 3 + \frac{- 4}{3} \cdot 9$

Factor $3$ out of $9$ .

$6 + 3 + \frac{- 4}{3} \cdot (3 (3))$

Cancel the common factor.

$6 + 3 + \frac{- 4}{3} \cdot (3 \cdot 3)$

Rewrite the expression.

$6 + 3 - 4 \cdot 3$

Multiply $- 4$ by $3$ .

$6 + 3 - 12$

Simplify by adding and subtracting.

Add $6$ and $3$ .

$9 - 12$

Subtract $12$ from $9$ .

$- 3$