The difference between two positive numbers is 4 and the difference between their cubes is 316. Find :

(i) their product

(ii) the sum of their squares.

If x² + $\dfrac{1}{\text{x}^2}$ = 3, the value of x - $\dfrac{1}{\text{x}}$ is:

1. 1

2. -1

3. $\pm 1$

4. 0

$\dfrac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204}$ is equal to:

1. 400

2. -8

3. 8

4. none of these

If a = x + $\dfrac{1}{\text{x}}$ and b = x - $\dfrac{1}{\text{x}}$, then a² - b² is:

1. x² + $\dfrac{1}{\text{x}^2}$

2. x² - $\dfrac{1}{\text{x}^2}$

3. 4

4. $2\Big(x^2 + \dfrac{1}{\text{x}^2}\Big)$