Find the continued product:

Solution:

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

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

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

$\Rightarrow\left(x^{2}\right)^{2}-\left(1^{2}\right)^{2}$   [according to the formula $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$

$\Rightarrow x^{4}-1 .$

Therefore, the product of $(x+1)(x-1)\left(x^{2}+1\right)$ is $x^{4}-1$.

(ii) $(x-3)(x+3)\left(x^{2}+9\right)$

$\Rightarrow\left((x)^{2}-(3)^{2}\right)\left(x^{2}+9\right)$   [according to the formula $a^{2}-b^{2}=(a+b)(a-b)$ ]

$\Rightarrow\left(x^{2}-9\right)\left(x^{2}+9\right)$

$\Rightarrow\left(x^{2}\right)^{2}-(9)^{2}$    [according to the formula $a^{2}-b^{2}=(a+b)(a-b)$ ]

$\Rightarrow x^{4}-81$

Therefore, the product of $(x-3)(x+3)\left(x^{2}+9\right)$ is $x^{4}-81$.

(iii) $(3 x-2 y)(3 x+2 y)\left(9 x^{2}+4 y^{2}\right)$

$\Rightarrow\left((3 x)^{2}-(2 y)^{2}\right)\left(9 x^{2}+4 y^{2}\right)$   [according to the formula $a^{2}-b^{2}=(a+b)(a-b)$ ]

$\Rightarrow\left(9 x^{2}-4 y^{2}\right)\left(9 x^{2}+4 y^{2}\right)$

$\Rightarrow\left(9 x^{2}\right)^{2}-\left(4 y^{2}\right)^{2}$   [according to the formula $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$

$\Rightarrow 81 x^{4}-16 y^{4} .$   

Therefore, the product of $(3 x-2 y)(3 x+2 y)\left(9 x^{2}+4 y^{2}\right)$ is $81 x^{4}-16 y^{4}$.

(iv) $(2 p+3)(2 p-3)\left(4 p^{2}+9\right)$

$\Rightarrow\left((2 p)^{2}-(3)^{2}\right)\left(4 p^{2}+9\right)$   $\left[\right.$ according to the formula $\left.a^{2}-b^{2}=(a+b)(a-b)\right]$

$\Rightarrow\left(4 p^{2}-9\right)\left(4 p^{2}+9\right)$

$\Rightarrow\left(4 p^{2}\right)^{2}-(9)^{2}$   [according to the formula $a^{2}-b^{2}=(a+b)(a-b)$ ]

$\Rightarrow 16 p^{4}-81$

Therefore, the product of $(2 p+3)(2 p-3)\left(4 p^{2}+9\right)$ is $16 p^{4}-81$.