# Let a^6 - b^6 is simplified to (a - b) (a^2 - ab + b^2) k , then k is?

## Let ${a}^{6}$ - ${b}^{6}$ is simplified to (a - b) (${a}^{2}$ - ab + ${b}^{2}$) k, then k is?

Mar 20, 2018

The solution is $k = \left(a + b\right) \left({a}^{2} + a b + {b}^{2}\right)$.

#### Explanation:

${a}^{6} - {b}^{6} = \left(a - b\right) \left({a}^{2} - a b + {b}^{2}\right) k$

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

Difference of squares factoring:

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

Difference of cubes factoring:

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

$\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a - b\right)}}} \left({a}^{2} + a b + {b}^{2}\right) \left({a}^{3} + {b}^{3}\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a - b\right)}}} \left({a}^{2} - a b + {b}^{2}\right) k$

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

Sum of cubes factoring:

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

$\left({a}^{2} + a b + {b}^{2}\right) \left(a + b\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({a}^{2} - a b + {b}^{2}\right)}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({a}^{2} - a b + {b}^{2}\right)}}} k$

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

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

That's the answer. Hope this helped!