Simplify the following expressions: 5d) 3/(2x3) - sqrt((8x2y3)/(2y)) 5f) 3/(2x3) - sqrt((32y8)/(9y2)) + sqrt((98y6)/(2))

For the first part:

$$\frac{54x^6y^3}{2x^3} = (54 \div 2) \cdot (x^6 \div x^3) \cdot y^3 = 27x^{6-3}y^3 = 27x^3y^3$$

For the second part:

$$\frac{8x^2y^3}{2y} = (8 \div 2) \cdot x^2 \cdot (y^3 \div y) = 4x^2y^{3-1} = 4x^2y^2$$

Now the expression is:

$$\sqrt[3]{27x^3y^3} - \sqrt{4x^2y^2}$$

For the first term, $\sqrt[3]{27x^3y^3}$:

$$\sqrt[3]{27} \cdot \sqrt[3]{x^3} \cdot \sqrt[3]{y^3} = 3 \cdot x \cdot y = 3xy$$

For the second term, $\sqrt{4x^2y^2}$:

$$\sqrt{4} \cdot \sqrt{x^2} \cdot \sqrt{y^2} = 2 \cdot x \cdot y = 2xy$$

$$3xy - 2xy = xy$$

The final answer is $\boxed{xy}$.

5f)
The expression is:

$$\sqrt[3]{\frac{16x^3y^9}{2x^3}} - \sqrt{\frac{32y^8}{9y^2}} + \sqrt{\frac{98y^6}{2}}$$

For the first part:

$$\frac{16x^3y^9}{2x^3} = (16 \div 2) \cdot (x^3 \div x^3) \cdot y^9 = 8 \cdot 1 \cdot y^9 = 8y^9$$

For the second part:

$$\frac{32y^8}{9y^2} = \frac{32}{9} \cdot (y^8 \div y^2) = \frac{32y^6}{9}$$

For the third part:

$$\frac{98y^6}{2} = (98 \div 2) \cdot y^6 = 49y^6$$

Now the expression is:

$$\sqrt[3]{8y^9} - \sqrt{\frac{32y^6}{9}} + \sqrt{49y^6}$$

For the first term, $\sqrt[3]{8y^9}$:

$$\sqrt[3]{8} \cdot \sqrt[3]{y^9} = 2 \cdot y^{(9 \div 3)} = 2y^3$$

For the second term, $\sqrt{\frac{32y^6}{9}}$:

$$\frac{\sqrt{32y^6}}{\sqrt{9}} = \frac{\sqrt{16 \cdot 2 \cdot y^6}}{3} = \frac{\sqrt{16} \cdot \sqrt{2} \cdot \sqrt{y^6}}{3} = \frac{4 \cdot \sqrt{2} \cdot y^{(6 \div 2)}}{3} = \frac{4y^3\sqrt{2}}{3}$$

For the third term, $\sqrt{49y^6}$:

$$\sqrt{49} \cdot \sqrt{y^6} = 7 \cdot y^{(6 \div 2)} = 7y^3$$

$$2y^3 - \frac{4y^3\sqrt{2}}{3} + 7y^3$$

Combine the terms that have $y^3$:

$$(2y^3 + 7y^3) - \frac{4y^3\sqrt{2}}{3} = 9y^3 - \frac{4y^3\sqrt{2}}{3}$$

To combine these, write $9y^3$ with a denominator of 3:

$$\frac{9 \cdot 3}{3}y^3 - \frac{4\sqrt{2}}{3}y^3 = \frac{27y^3}{3} - \frac{4\sqrt{2}y^3}{3}$$ $$= \frac{27y^3 - 4\sqrt{2}y^3}{3} = \frac{y^3(27 - 4\sqrt{2})}{3}$$

The final answer is $\boxed{\frac{y^3(27 - 4\sqrt{2})}{3}}$.