# How do you simplify (7x-1)(49x^2+7x+1)(343x^3+1)?

Jun 14, 2015

$\left(7 x - 1\right) \left(49 {x}^{2} + 7 x + 1\right) \left(343 {x}^{3} + 1\right) = {\left(7 x\right)}^{6} - 1$

#### Explanation:

${\left(7 x\right)}^{6} - 1$

$= {\left({\left(7 x\right)}^{3}\right)}^{2} - {1}^{2}$

$= \left({\left(7 x\right)}^{3} - 1\right) \left({\left(7 x\right)}^{3} + 1\right)$

$= \left({\left(7 x\right)}^{3} - {1}^{3}\right) \left({\left(7 x\right)}^{3} + 1\right)$

$= \left(7 x - 1\right) \left({\left(7 x\right)}^{2} + 7 x + 1\right) \left({\left(7 x\right)}^{3} + 1\right)$

$= \left(7 x - 1\right) \left(49 {x}^{2} + 7 x + 1\right) \left(343 {x}^{3} + 1\right)$

using the difference of square identity:

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

and the difference of cubes identity:

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