(i) (a/2b+2b/a)^2-(2b/a-a/2b)^2 (ii) (4a+3b)^2-(4a-3b)^2+48ab (iii) (x-2)(x-3)(x+4) [S.Chand ICSE Class 9 Fundamentals of Mathematics Chapter 3 Exercise 3.1 Expansion and Factorisation Solution ]
(i) (a/2b+2b/a)^2-(2b/a-a/2b)^2
$\left(\frac{a}{2b} + \frac{2b}{a}\right)^2 – \left(\frac{2b}{a} – \frac{a}{2b}\right)^2$
= $\left(\frac{a}{2b} + \frac{2b}{a} + \frac{2b}{a} – \frac{a}{2b}\right)\left(\frac{a}{2b} + \frac{2b}{a} – \frac{2b}{a} + \frac{a}{2b}\right)$ [∵ a2-b2=(a+b)(a-b)]
= $\frac{4b}{a} \times \frac{2a}{2b}$
= $\frac{4ab}{ab}$
= 4
(ii) (4a+3b)2-(4a-3b)2+48ab
= {(4a+3b)+(4a-3b)}{(4a+3b)-(4a-3b)}+48ab
=(4a+3b+4a-3b)(4a+3b-4a+3b)+48ab
=8a x 6b + 48ab
= 48ab+48ab
= 96ab
(iii) (x-2)(x-3)(x+4)
={(x-2)(x-3)}(x+4)
={x(x-3)-2(x-3)}(x+4)
=(x2-3x-2x+6)(x+4)
=(x2-5x+6)(x+4)
=x(x2-5x+6)+4(x2-5x+6)
=x3-5x2+6x+4x2-20x+24
=x3-x2-14x+24