# How do you factor 64t^6-1?

Oct 8, 2016

$\left(2 t - 1\right) \left(2 t + 1\right) \left(16 {t}^{4} + 4 {t}^{2} + 1\right)$

#### Explanation:

This is a $\textcolor{b l u e}{\text{difference of cubes}}$ and in general, factorises as follows.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots \left(A\right)$

Note that.

${\left(4 {t}^{2}\right)}^{3} = 64 {t}^{6} \text{ and } {\left(1\right)}^{3} = 1$

$\Rightarrow a = 4 {t}^{2} \text{ and } b = 1$

substitute these values for a and b into (A)

$\left(4 {t}^{2} - 1\right) \left(16 {t}^{4} + 4 {t}^{2} + 1\right)$

We now have a factor, $\left(4 {t}^{2} - 1\right)$ which is a $\textcolor{b l u e}{\text{difference of squares}}$ and factorises, in general as.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots \left(B\right)$

${\left(2 t\right)}^{2} = 4 {t}^{2} \text{ and } {\left(1\right)}^{2} = 1$

$\Rightarrow a = 2 t \text{ and } b = 1$

substitute these values for a and b into (B)

$\left(2 t - 1\right) \left(2 t + 1\right)$

Pulling all this together.

$64 {t}^{6} - 1 = \left(4 {t}^{2} - 1\right) \left(16 {t}^{4} + 4 {t}^{2} + 1\right)$

$= \left(2 t - 1\right) \left(2 t + 1\right) \left(16 {t}^{4} + 4 {t}^{2} + 1\right)$

The factor $16 {t}^{4} + 4 {t}^{2} + 1 \text{ may be factorised as follows}$

This is a polynomial of degree 4, with no odd degree terms.

${a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4} = \left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right)$

here a = 2t , b = 1 and k = 1

$\Rightarrow 16 {t}^{4} + 4 {t}^{2} + 1 = \left(4 {t}^{2} - 2 t + 1\right) \left(4 {t}^{2} + 2 t + 1\right)$

The whole expression now factorises to.

$64 {t}^{6} - 1 = \left(2 t - 1\right) \left(2 t + 1\right) \left(4 {t}^{2} - 2 t + 1\right) \left(4 {t}^{2} + 2 t + 1\right)$

I thank George C, for his input to the factorising of the quartic.

Oct 28, 2016

Use the difference of squares, difference of cubes and sum of cubes identities to find:

$64 {t}^{6} - 1 = \left(2 t - 1\right) \left(4 {t}^{2} + 2 t + 1\right) \left(2 t + 1\right) \left(4 {t}^{2} - 2 t + 1\right)$

#### Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

The sum of cubes identity can be written:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Hence we find:

$64 {t}^{6} - 1 = {\left(8 {t}^{3}\right)}^{2} - {1}^{2}$

$\textcolor{w h i t e}{64 {t}^{6} - 1} = \left(8 {t}^{3} - 1\right) \left(8 {t}^{3} + 1\right)$

$\textcolor{w h i t e}{64 {t}^{6} - 1} = \left({\left(2 t\right)}^{3} - {1}^{3}\right) \left({\left(2 t\right)}^{3} + {1}^{3}\right)$

$\textcolor{w h i t e}{64 {t}^{6} - 1} = \left(2 t - 1\right) \left({\left(2 t\right)}^{2} + \left(2 t\right) + 1\right) \left(\left(2 t\right) + 1\right) \left({\left(2 t\right)}^{2} - \left(2 t\right) + 1\right)$

$\textcolor{w h i t e}{64 {t}^{6} - 1} = \left(2 t - 1\right) \left(4 {t}^{2} + 2 t + 1\right) \left(2 t + 1\right) \left(4 {t}^{2} - 2 t + 1\right)$