Find each of the following products:

Solution:

(i) We have:

$(x-4)(x-4)$

$=(x-4)^{2}$

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

$=x^{2}-8 x+16$

(ii) We have:

$(2 x-3 y)(2 x-3 y)$

$=(2 x-3 y)^{2}$

$=(2 x)^{2}-2 \times 2 x \times 3 y+(3 y)^{2}$   $\left[\right.$ using $\left.(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

$=4 x^{2}-12 x y+9 y^{2}$

(iii) $\left(\frac{3}{4} x-\frac{5}{6} y\right)\left(\frac{3}{4} x-\frac{5}{6} y\right)$

$=\left(\frac{3}{4} x-\frac{5}{6} y\right)^{2}$

$=\left(\frac{3}{4} x\right)^{2}-2 \times \frac{3}{4} x \times \frac{5}{6} y+\left(\frac{5}{6} y\right)^{2}$  $\left[\right.$ using $\left.(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

$=\frac{9}{16} x^{2}-\frac{15}{12} x y+\frac{25}{36} y^{2}$

(iv) We have:

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

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

$=(x)^{2}-2 \times x \times \frac{3}{x}+\left(\frac{3}{x}\right)^{2}$   $\left[\right.$ using $\left.(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

$=x^{2}-6+\frac{9}{x^{2}}$

(v) We have:

$\left(\frac{1}{3} x^{2}-9\right)\left(\frac{1}{3} x^{2}-9\right)$

$=\left(\frac{1}{3} x^{2}-9\right)^{2}$

$=\left(\frac{1}{3} x^{2}\right)^{2}-2 \times \frac{1}{3} x^{2} \times 9+(9)^{2}$   $\left[\operatorname{using}(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

$=\frac{1}{9} x^{4}-6 x^{2}+81$

(vi) We have:

$\left(\frac{1}{2} y^{2}-\frac{1}{3} y\right)\left(\frac{1}{2} y^{2}-\frac{1}{3} y\right)$

$=\left(\frac{1}{2} y^{2}-\frac{1}{3} y\right)^{2}$

$=\left(\frac{1}{2} y^{2}\right)^{2}-2 \times \frac{1}{2} y^{2} \times \frac{1}{3} y+\left(\frac{1}{3} y\right)^{2}$    $\left[\operatorname{using}(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

$=\frac{1}{4} y^{4}-\frac{1}{3} y^{3}+\frac{1}{9} y^{2}$