Find the following products using identities:

(A) $(-3x + 4)^2$

(B) $(2s + 7)(2s - 7)$

(C) $\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)$

(D) $(2n + 7)(2n - 7)$

(E) $(s - 2t)(s^2 + 2st + 4t^2)$

(F) $\left(\frac{1}{2r} - 4r\right)^2$

(G) $(-3m + 4k - l)^2$

(H) $\left(x - \frac{y}{3}\right)^3$

(I) $\left(\frac{7k}{2} - \frac{2m}{3}\right)^3$

(A) $(-3x + 4)^2 = (-3x)^2 + 2(-3x)(4) + 16$

$$= 9x^2 - 24x + 16$$

(B) $(2s + 7)(2s - 7) = (2s)^2 - 7^2$

$$= 4s^2 - 49$$

(C) $\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right) = (p^2)^2 - \left(\frac{1}{2}\right)^2$

$$= p^4 - \frac{1}{4}$$

(D) $(2n + 7)(2n - 7) = 4n^2 - 49$

(E) $(s - 2t)(s^2 + 2st + 4t^2) = s^3 - (2t)^3$

$$= s^3 - 8t^3$$

$$[\text{Using } a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$$

(F) $\left(\frac{1}{2r} - 4r\right)^2 = \left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2$

$$= \frac{1}{4r^2} - 4 + 16r^2$$

(G) $(-3m + 4k - l)^2$

$$= 9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6ml$$

Apply $(a + b + c)^2$ formula with $a = -3m$ ,

$$b = 4k, c = -l:$$

$$a^2 + b^2 + c^2 = 9m^2 + 16k^2 + l^2$$

$$2ab = 2(-3m)(4k) = -24mk$$

$$2bc = 2(4k)(-l) = -8kl$$

$$2ca = 2(-l)(-3m) = 6lm$$

$$\text{Sum: } 9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm$$

(H) $\left(x - \frac{y}{3}\right)^3 = x^3 - 3x^2\left(\frac{y}{3}\right) + 3x\left(\frac{y}{3}\right)^2 - \left(\frac{y}{3}\right)^3$

$$= x^3 - x^2y + \frac{xy^2}{3} - \frac{y^3}{27}$$

(I) $\left(\frac{7k}{2} - \frac{2m}{3}\right)^3$

Using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=\frac{7k^{}}{2}$, $b=\frac{2m^{}}{3}$:

$$\text{So, } \left(\frac{7k}{2} - \frac{2m}{3}\right)^3 =$$

$$\frac{343k^3}{8} - \frac{49k^2m}{2} + \frac{14km^2}{3} - \frac{8m^3}{27}$$