Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(A) $\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

(B) $\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

(C) $\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$

(D) $\frac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2}$

(E) $\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$

(F) $\frac{p^4 - 16}{p^2 - 4p + 4}$                          [Page No. 87]

12. (A) First factorising the numerator.

$$\begin{aligned} 3p^2 - 3pq - 18q^2 \\ &= 3(p^2 - pq - 6q^2) \\ &= 3[p^2 - 3pq + 2pq - 6q^2] \\ &= 3[p(p - 3q) + 2q(p - 3q)] \\ &= 3(p - 3q)(p + 2q) \end{aligned}$$

Now factorising the denominator

$$\begin{aligned} p^2 + 3pq - 10q^2 \\ &= p^2 + 5pq - 2pq - 10q^2 \\ &= p(p + 5q) - 2q(p + 5q) \\ &= (p + 5q)(p - 2q) \end{aligned}$$

$$\text{So, } \frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}$$

No common factor cancels.

(B) Factorising the numerator using identity:

$$\begin{aligned} n^2 - 3n^2m + 3nm^2 - m^3 \\ &= (n - m)^2 \\ &\quad [\text{Using } a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3] \end{aligned}$$

Factorising the denominator:

$$\begin{aligned} 5m^2 - 10mn + 5n^2 \\ &= 5(m^2 - 2mn + n^2) \\ &= 5(m - n)^2 \quad [\text{Using } a^2 - 2ab + b^2 = (a - b)^2] \end{aligned}$$

$$\text{Now, } \frac{(n - m)^3}{[5(m - n)^2]}$$

$$=\frac{(m-n)^3}{\left\lbrack5\left(m-n\right)^2\right\rbrack}=-(m-n)^5$$

(C) Factorising the numerator:

$$\begin{aligned}w)^3-v^3+x^3+3wvx & \\ & =w^3+(-v)^3+x^3-3w(-v)x\\ & =(w-v+x)(w^2+v^2+x^2+wv+vx-wx)\\ & \quad[\text{Using }a^3+b^3+c^3+3abc=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)]\\ & \end{aligned}$$

Now denominator:

$$\begin{aligned} w^2 + v^2 + x^2 - 2wv - 2vx + 2wx \\ &= w^2 + (-v)^2 + x^2 + 2w(-v) + 2(-v)x + 2xw \\ &= (w - v + x) \\ &\quad [\text{Using } a^2 + b^2 + c^2 + 2ab + 2bc \\ &\quad + 2ca = (a + b + c)^2] \end{aligned}$$

$$\text{Therefore, } \frac{(w^3 - v^3 + x^3 + 3wx)}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$$

$$= \frac{(w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)}{(w - v + x)^2}$$

$$= \frac{(w^2 + x^2 + v^2 - wx + vx + wv)}{w + x - v}$$[Canceling the common factor]

(D) Factor numerator:

$$\begin{aligned}4y^2-20yz+25z^2 & \\ & =(2y)^2-2(2y)(5z)+(5z)^2\\ & =(2y-5z)^2\quad[\text{Using }a^2-2ab+b^2=(a-b)^2]\\ & \quad\text{Factorising the denominator:}\\ 25z^2-4y^2 & =(5z-2y)(5z+2y)\\ & \quad[\text{Using }a^2-b^2=(a-b)(a+b)]\\ & \text{So, }\frac{(4y^2 - 20yz + 25z^2)}{(25z^2 - 4y^2)}\\ & =\frac{(5z - 2y)^2}{(5z - 2y)(5z + 2y)}\\ & \quad[\text{Since }(2y-5z)^2=(-1)^2(5z-2y)^2=(5z-2y)^2]\\ & \\ & =\frac{(5z - 2y)}{(5z + 2y)}\end{aligned}$$

$$(E) \frac{(x^2 + x - 6)(x^2 - 7x + 12)}{[(x^2 - 6x + 8)(x^2 - 9)]}$$

$$x^2 + x - 6 = (x + 3)(x - 2)$$

$$x^2 - 7x + 12 = (x - 3)(x - 4)$$

$$x^2 - 6x + 8 = (x - 2)(x - 4)$$

$$x^2 - 9 = (x - 3)(x + 3)$$

Hence, given expression can be written as:

$$\frac{(x + 3)(x - 2)(x - 3)(x - 4)}{(x - 2)(x - 4)(x - 3)(x + 3)} = 1$$

$$(F) \frac{p^4 - 16}{p^2 - 4p + 4}$$

$$\text{Numerator} = (p^2 - 4)(p^2 + 4)$$

$$= (p - 2)(p + 2)(p^2 + 4)$$

$$\text{Denominator} = (p - 2)^2$$

Hence, given expression can be written as:

$$\frac{(p + 2)(p^2 + 4)}{p - 2}$$

Caution
Students should fully factorise the numerator and denominator before cancelling in a rational expression, and should cancel only common factors, never individual terms taken out of a sum.