Simplify each of the following products:

Solution:

(i) We have,

$\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\frac{1}{4} a^{2}+9 b^{2}\right)$

$\left.\left.\Rightarrow\left[\left(\frac{1}{2} a\right)^{2}-(3 b)^{2}\right)\right]\left[\frac{1}{4} a^{2}+9 b^{2}\right] \quad\left[\therefore(a+b)(a-b)=a^{2}-b^{2}\right)\right]$

$\left.\left.\Rightarrow\left[\frac{1}{4} a^{2}-9 b^{2}\right)\right]\left[\frac{1}{4} a^{2}+9 b^{2}\right] \quad\left[\therefore(a b)^{2}=a^{2} b^{2}\right)\right]$

$\left.=\left[\left(\frac{1}{4} a^{2}\right)^{2}-\left(9 b^{2}\right)^{2}\right] \quad\left[\therefore(a+b)(a-b)=a^{2}-b^{2}\right)\right]$

$=\frac{1}{16} a^{4}-81 b^{4}$

$\therefore\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\frac{1}{4} a^{2}+9 b^{2}\right)=\frac{1}{16} a^{4}-81 b^{4}$

(ii) We have,

We have,

$(m+n / 7)^{3}(m-n / 7)$

$=(m+n / 7)(m+n / 7)(m+n / 7)(m-n / 7)$

$=(m+n / 7)^{2}\left(m^{2}-(n / 7)^{2}\right)\left[\therefore(a+b)(a+b)=(a+b)^{2}\right.$ and $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

$=\left(m+n^{7}\right)^{2}\left(m^{2}-n^{2} / 49\right)$

$\therefore(m+n / 7)^{3}(m-n / 7)=(m+n / 7)^{2}\left(m^{2}-n^{2} / 49\right)$

(iii) We have,

$(x / 2-2 / 5)(2 / 5-x / 2)-x^{2}+2 x$

$\Rightarrow-(2 / 5-x / 2)(2 / 5-x / 2)-x^{2}+2 x$

$\Rightarrow-(2 / 5-x / 2)^{2}-x^{2}+2 x \quad\left[\because(a-b)(a-b)=(a-b)^{2}\right]$

$\Rightarrow-\left[(2 / 5)^{2}+(x / 2)^{2}-2(2 / 5)(x / 2)\right]-x^{2}+2 x$

$\Rightarrow-\left(4 / 25+x^{2} / 4-2 x / 5\right)-x^{2}+2 x$

$\Rightarrow-x^{2} / 4-x^{2}+2 x / 5+2 x-4 / 25$

$\Rightarrow-5 x^{2} / 4+12 x / 5-4 / 25$

$\therefore(x / 2-2 / 5)(2 / 5-x / 2)-x^{2}+2 x=-5 x^{2} / 4+12 x / 5-4 / 25$

(iv) We have,

$\left(x^{2}+x-2\right)\left(x^{2}-x+2\right)$

$\left[(x)^{2}+(x-2)\right]\left(x^{2}-(x+2)\right]$

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

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

$\Rightarrow x 4-x^{2}+4 x-4$

$\therefore\left(x^{2}+x-2\right)\left(x^{2}-x+2\right)=x^{4}-x^{2}+4 x-4$

(v) we have,

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

$\Rightarrow x\left(x^{2}-3 x-1\right)\left(x^{2}-3 x+1\right)$

$\Rightarrow x\left[\left(x^{2}-3 x\right)^{2}-(1)^{2}\right] \quad\left[\therefore(a+b)(a-b)=a^{2}-b^{2}\right]$

$\Rightarrow x\left[\left(x^{2}\right)^{2}+(-3 x)^{2}-2(3 x)\left(x^{2}\right)-1\right]$

$\Rightarrow x\left[x^{4}+9 x^{2}-6 x^{3}-1\right]$

$\Rightarrow x^{5}-6 x^{4}+9 x^{3}-x$

$\therefore\left(x^{3}-3 x-x\right)\left(x^{2}-3 x+1\right)=x^{5}-6 x^{4}+9 x^{3}-x$

(vi) We have,

$\left(2 x^{4}-4 x^{2}+1\right)\left(2 x^{4}-4 x^{2}-1\right)$

$\Rightarrow\left[\left(2 x^{4}-4 x^{2}\right)^{2}-(1)^{2}\right]\left[\therefore(a+b)(a-b)=a^{2}-b^{2}\right]$

$\Rightarrow\left[\left(2 x^{4}\right)^{2}+\left(4 x^{2}\right)^{2}-2\left(2 x^{4}\right)\left(4 x^{2}\right)-1\right]$

$\Rightarrow 4 x^{8}-16 x^{6}+16 x^{4}-1\left[\therefore(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

$\therefore\left(2 x^{4}-4 x^{2}+1\right)\left(2 x^{4}-4 x^{2}-1\right)=4 x^{8}-16 x^{6}+16 x^{4}-1$