Factor the following algebraic expressions:

(A) $4y^2 + 1 + \frac{1}{16y^2}$

(B) $9m^2 - \frac{1}{25n^2}$

(C) $27b^3 - \frac{1}{64b^3}$

(D) $x^2 + \frac{5x}{6} + \frac{1}{6}$

(E) $27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}$

(F) $64y^3 + \frac{z^3}{125}$

(G) $p^3 + 27q^3 + r^3 - 9pqr$

(H) $9m^2 - 12m + 4$

(I) $9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz$

** (J) $4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz$

(This question is incorrect in NCERT Book. The question here is modified accordingly.)

(K) $27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}$

(A) $4y^2 + 1 + \frac{1}{16y^2} = (2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2$

$$= \left(2y + \frac{1}{4y}\right)^2$$

(B) $9m^2 - \frac{1}{25n^2} = (3m)^2 - \left(\frac{1}{5n}\right)^2$

$$= \left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right)$$

(C) $27b^3 - \frac{1}{64b^3} = (3b)^3 - \left(\frac{1}{4b}\right)^3$

$$= \left(3b - \frac{1}{4b}\right)\left[(3b)^2 + (3b)\left(\frac{1}{4b}\right) + \frac{1}{4b^2}\right]$$

(D) $x^2 + \frac{5x}{6} + \frac{1}{6}$

Multiply/divide appropriately;

$$\text{we need } a + b = \frac{5}{6},\ ab = \frac{1}{6}:$$

$$a = \frac{1}{2},\ b = \frac{1}{3}.$$

$$x^2 + \frac{5x}{6} + \frac{1}{6} = \left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)$$

$$(E)\ 27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}$$

$$= (3u)^3 - \left(\frac{1}{5}\right)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2$$

$$= (3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3$$

$$= \left(3u - \frac{1}{5}\right)^3$$

$$(F)\ 64y^3 + \frac{z^3}{125} = (4y)^3 + \left(\frac{z}{5}\right)^3$$

$$= \left(4y + \frac{z}{5}\right)\left[(4y)^2 - (4y)\left(\frac{z}{5}\right) + \left(\frac{z}{5}\right)^2\right]$$

$$= \left(4y + \frac{z}{5}\right)\left(16y^2 - \frac{4yz}{5} + \frac{z^2}{25}\right)$$

$$(G)\ p^3 + 27q^3 + r^3 - 9pqr$$

$$= p^3 + (3q)^3 + r^3 - 3p\cdot(3q)r$$

Using $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

$$\text{Here, } a = p,\ b = 3q,\ c = r:$$

Hence, given expression in factorised form:

$$(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)$$

$$(H)\ 9m^2 - 12m + 4 = (3m)^2 - 2(3m)(2) + (2)^2$$

$$= (3m - 2)^2$$

$$(I)\ 9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz$$

$$= \left(\frac{1}{3}\right)[27x^3 - 8y^3 + z^3 + 18xyz]$$

$$= \left(\frac{1}{3}\right)[(3x)^3 + (-2y)^3 + z^3 - 3(3x)(-2y)z]$$

$$=\frac13(3x-2y+z)\left[(3x)^2+(-2y)^2+z^2-(3x)(-2y)-(-2y)(z)-(z)(3x)\right]$$

$$=\frac{1}{3}(3x-2y+z)\left[9x^2+4y^2+z^2+6xy+2yz-3zx\right]$$

$$(J)\ 4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz$$

$=(2x)^2+(3y)^2+(6z)^2+2(2x)(3y)+2(3y)(6z)+2(2x)(6z)$

$=(2x+3y+6z)^2$
 [Using $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2$]

$$(K)\ 27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}$$

$$= (3u)^3 - \left(\frac{1}{6}\right)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2$$

$$= (3u)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^3$$

$$=\left(3u-\frac{1}{6}\right)^3$$