Answer
$a.\quad(2-3n)(2+3n)(4+9n^{2})$
$b.\quad 16$ and $81n^{4}$ are both squares of square terms
$c.\quad$ (sample answer)$\quad 16x^{4}-1$
Work Step by Step
a.
$16-81n^{4}=$
...write as the difference of two squares.
...($a=4,\ b=9n^{2})$
$=(4-9n^{2})(4+9n^{2})$
...the expression in the first parentheses can be written as a difference of two squares.
...($a=2,b=3n$)
$=(2-3n)(2+3n)(4+9n^{2})$
b.
$16$ and $81n^{4}$ are both squares of square terms.
c.
Take for example, $2x$ and 1, square them twice
$[(2x)^{2}]^{2} =[4x^{2}]^{2}=16x^{4},$
$[1^{2}]^{2}=1$
$16x^{4}-1$ can be factored twice:
$=(4x^{2}-1)(4x^{2}+1)$
$=(2x+1)(2x-1)(4x^{2}+1)$
