Find each of the following products:
(i) `(x - 4)(x - 4)` (ii) `(2x - 3y)(2x - 3y)` (iii) `((3)/(4) x - (5)/(6) y) ((3)/(4)x - (5)/(6) y)`
(iv) `(x - (3)/(x)) (x - (3)/(x))` (v) `((1)/(3) x^(2) - 9) ((1)/(3) x^(2) - 9)` (vi) `((1)/(2) y^(2) - (1)/(3) y) ((1)/(2) y^(2) - (1)/(3) y)`

Answer

To find the products of the given expressions, we will use the identity for the square of a binomial, which states that:

\[ (a - b)^2 = a^2 - 2ab + b^2 \]

Now, let's solve each part step by step.

### (i) \((x - 4)(x - 4)\)

1. Recognize that this is \((x - 4)^2\).
2. Apply the identity:
- Here, \(a = x\) and \(b = 4\).
- So, \((x - 4)^2 = x^2 - 2(4)(x) + 4^2\).
3. Calculate:
- \(x^2 - 8x + 16\).

**Final Answer:** \(x^2 - 8x + 16\)

### (ii) \((2x - 3y)(2x - 3y)\)

1. Recognize that this is \((2x - 3y)^2\).
2. Apply the identity:
- Here, \(a = 2x\) and \(b = 3y\).
- So, \((2x - 3y)^2 = (2x)^2 - 2(2x)(3y) + (3y)^2\).
3. Calculate:
- \(4x^2 - 12xy + 9y^2\).

**Final Answer:** \(4x^2 - 12xy + 9y^2\)

### (iii) \(\left(\frac{3}{4}x - \frac{5}{6}y\right)\left(\frac{3}{4}x - \frac{5}{6}y\right)\)

1. Recognize that this is \(\left(\frac{3}{4}x - \frac{5}{6}y\right)^2\).
2. Apply the identity:
- Here, \(a = \frac{3}{4}x\) and \(b = \frac{5}{6}y\).
- So, \(\left(\frac{3}{4}x - \frac{5}{6}y\right)^2 = \left(\frac{3}{4}x\right)^2 - 2\left(\frac{3}{4}x\right)\left(\frac{5}{6}y\right) + \left(\frac{5}{6}y\right)^2\).
3. Calculate:
- \(\frac{9}{16}x^2 - \frac{15}{12}xy + \frac{25}{36}y^2\).

**Final Answer:** \(\frac{9}{16}x^2 - \frac{15}{12}xy + \frac{25}{36}y^2\)

### (iv) \(\left(x - \frac{3}{x}\right)\left(x - \frac{3}{x}\right)\)

1. Recognize that this is \(\left(x - \frac{3}{x}\right)^2\).
2. Apply the identity:
- Here, \(a = x\) and \(b = \frac{3}{x}\).
- So, \(\left(x - \frac{3}{x}\right)^2 = x^2 - 2\left(x\right)\left(\frac{3}{x}\right) + \left(\frac{3}{x}\right)^2\).
3. Calculate:
- \(x^2 - 6 + \frac{9}{x^2}\).

**Final Answer:** \(x^2 - 6 + \frac{9}{x^2}\)

### (v) \(\left(\frac{1}{3}x^2 - 9\right)\left(\frac{1}{3}x^2 - 9\right)\)

1. Recognize that this is \(\left(\frac{1}{3}x^2 - 9\right)^2\).
2. Apply the identity:
- Here, \(a = \frac{1}{3}x^2\) and \(b = 9\).
- So, \(\left(\frac{1}{3}x^2 - 9\right)^2 = \left(\frac{1}{3}x^2\right)^2 - 2\left(\frac{1}{3}x^2\right)(9) + 9^2\).
3. Calculate:
- \(\frac{1}{9}x^4 - 6x^2 + 81\).

**Final Answer:** \(\frac{1}{9}x^4 - 6x^2 + 81\)

### (vi) \(\left(\frac{1}{2}y^2 - \frac{1}{3}y\right)\left(\frac{1}{2}y^2 - \frac{1}{3}y\right)\)

1. Recognize that this is \(\left(\frac{1}{2}y^2 - \frac{1}{3}y\right)^2\).
2. Apply the identity:
- Here, \(a = \frac{1}{2}y^2\) and \(b = \frac{1}{3}y\).
- So, \(\left(\frac{1}{2}y^2 - \frac{1}{3}y\right)^2 = \left(\frac{1}{2}y^2\right)^2 - 2\left(\frac{1}{2}y^2\right)\left(\frac{1}{3}y\right) + \left(\frac{1}{3}y\right)^2\).
3. Calculate:
- \(\frac{1}{4}y^4 - \frac{1}{3}y^3 + \frac{1}{9}y^2\).

**Final Answer:** \(\frac{1}{4}y^4 - \frac{1}{3}y^3 + \frac{1}{9}y^2\)

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