# Factor completely?

## $4 {x}^{16} {y}^{16} {z}^{3} - 4 {x}^{16} {z}^{3} - 4 {y}^{16} {z}^{3} + 4 {z}^{3}$

Feb 20, 2018

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({x}^{4} + 1\right) \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right) \left({y}^{8} + 1\right) \left({y}^{4} + 1\right) \left({y}^{2} + 1\right) \left(y + 1\right) \left(y - 1\right)$

#### Explanation:

$4 {x}^{16} {y}^{16} {z}^{3} - 4 {x}^{16} {z}^{3} - 4 {y}^{16} {z}^{3} + 4 {z}^{3}$

let ${x}^{16} = a$

let ${y}^{16} = b$

let ${z}^{3} = c$

So given expression = $4 a b c - 4 a c - 4 b c + 4 c$

$\implies 4 a c \left(b - 1\right) - 4 c \left(b - 1\right)$ ----(group first two and last two terms and take the common factor outside the bracket)

$\implies \left(4 a c - 4 c\right) \left(b - 1\right)$

$\implies 4 c \left(a - 1\right) \left(b - 1\right)$

Now substitute the original values of a,b, and c:

$\implies 4 {z}^{3} \left({x}^{16} - 1\right) \left({y}^{16} - 1\right)$

Use identity: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

=> 4 z^3 [(x^8)^2 - 1^2)(y^8)^2 -1^2)]

$\implies 4 {z}^{3} \left[\left({x}^{8} - 1\right) \left({x}^{8} + 1\right)\right] \left[\left({y}^{8} - 1\right) \left({y}^{8} + 1\right)\right]$

$\implies 4 {z}^{3} \left[\left({x}^{8} + 1\right) \left({y}^{8} + 1\right)\right] \left[\left({x}^{8} - 1\right) \left({y}^{8} - 1\right)\right]$

=> 4z^3 (x^8+1)(y^8+1) [(x^4)^2 -(1)^2)((y^4)^2 -(1)^2)]

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left[\left({x}^{4} + 1\right) \left({x}^{4} - 1\right) \left({y}^{4} + 1\right) \left({y}^{4} - 1\right)\right]$

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left[\left({x}^{4} - 1\right) \left({y}^{4} - 1\right)\right]$

=> 4z^3 (x^8+1)(y^8+1)(x^4+1)(y^4+1)[(x^2)^2-1^2)(y^2)^2-1^2)]

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left[\left({x}^{2} + 1\right) \left({x}^{2} - 1\right) \left({y}^{2} + 1\right) \left({y}^{2} - 1\right)\right]$

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left({x}^{2} + 1\right) \left({y}^{2} + 1\right) \left[\left({x}^{2} - 1\right) \left({y}^{2} - 1\right)\right]$

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left({x}^{2} + 1\right) \left({y}^{2} + 1\right) \left[\left(x - 1\right) \left(x + 1\right) \left(y - 1\right) \left(y + 1\right)\right]$

$\implies 4 {z}^{3} \left({x}^{8} + 1\right) \left({x}^{4} + 1\right) \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right) \left({y}^{8} + 1\right) \left({y}^{4} + 1\right) \left({y}^{2} + 1\right) \left(y + 1\right) \left(y - 1\right)$

Feb 20, 2018

$4 {z}^{3} \left({x}^{8} + 1\right) \left({x}^{4} + 1\right) \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right) \left({y}^{8} + 1\right) \left({y}^{4} + 1\right) \left({y}^{2} + 1\right) \left(y + 1\right) \left(y - 1\right)$.

#### Explanation:

$\textcolor{red}{4} {x}^{16} {y}^{16} \textcolor{red}{{z}^{3}} - \textcolor{red}{4} {x}^{16} \textcolor{red}{{z}^{3}} - \textcolor{red}{4} {y}^{16} \textcolor{red}{{z}^{3}} + \textcolor{red}{4 {z}^{3}}$,

$= \textcolor{red}{4 {z}^{3}} \left({x}^{16} {y}^{16} - {x}^{16} - {y}^{16} + 1\right)$,

$= 4 {z}^{3} \left\{{x}^{16} \left({y}^{16} - 1\right) - 1 \left({y}^{16} - 1\right)\right\}$,

$= 4 {z}^{3} \left\{\left({y}^{16} - 1\right) \left({x}^{16} - 1\right)\right\}$,

=4z^3[{(y^8)^2-1^2}{(x^8)^2-1^2},

$= 4 {z}^{3} \left\{\left({y}^{8} + 1\right) \left({y}^{8} - 1\right)\right\} \left\{\left({x}^{8} + 1\right) \left({x}^{8} - 1\right)\right\}$,

$= 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left\{{\left({x}^{4}\right)}^{2} - {1}^{2}\right\} \left\{{\left({y}^{4}\right)}^{2} - {1}^{2}\right\}$,

$= 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left\{\left({x}^{4} + 1\right) \left({x}^{4} - 1\right)\right\} \left\{\left({y}^{4} + 1\right) \left({y}^{4} - 1\right)\right\}$,

$= 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left\{{\left({x}^{2}\right)}^{2} - {1}^{2}\right\} \left\{{\left({y}^{2}\right)}^{2} - {1}^{2}\right\}$,

$= 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left\{\left({x}^{2} + 1\right) \left({x}^{2} - 1\right)\right\} \left\{\left({y}^{2} + 1\right) \left({y}^{2} - 1\right)\right\}$,

$= 4 {z}^{3} \left({x}^{8} + 1\right) \left({y}^{8} + 1\right) \left({x}^{4} + 1\right) \left({y}^{4} + 1\right) \left({x}^{2} + 1\right) \left({y}^{2} + 1\right) \left\{\left(x + 1\right) \left(x - 1\right)\right\} \left\{\left(y + 1\right) \left(y - 1\right)\right\}$,

$= 4 {z}^{3} \left({x}^{8} + 1\right) \left({x}^{4} + 1\right) \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right) \left({y}^{8} + 1\right) \left({y}^{4} + 1\right) \left({y}^{2} + 1\right) \left(y + 1\right) \left(y - 1\right)$.