Find each of the following products:

Solution:

(i) We have:

$(x+3)(x-3)$

$=x^{2}-9$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(ii) We have:

$(2 x+5)(2 x-5)$

$=4 x^{2}-25$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(iii) We have:

$(8+x)(8-x)$

$=64-x^{2}$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(iv) We have:

$(7 x+11 y)(7 x-11 y)$

$=49 x^{2}-121 y^{2}$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(v) We have:

$\left(5 x^{2}+\frac{3}{4} y^{2}\right)\left(5 x^{2}-\frac{3}{4} y^{2}\right)

$=25 x^{4}-\frac{9}{16} y^{4}$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(vi) We have:

$\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$

$=x^{2}-\frac{1}{x^{2}}$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(vii) We have:

$\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)

$=x^{2}-\frac{1}{x^{2}}$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

(viii) We have:

$\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}-\frac{1}{y}\right)$

$=\frac{1}{x^{2}}-\frac{1}{y^{2}}$$\quad\left[\operatorname{using}(a+b)(a-b)=a^{2}-b^{2}\right]$

(ix) We have:

$\left(2 a+\frac{3}{b}\right)\left(2 a-\frac{3}{b}\right)$

$=4 a^{2}-\frac{9}{b^{2}}$$\quad\left[\right.$ using $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$