# How do you factor 8(5x + 7)^3 + 27(x-9)^3?

Dec 31, 2015

Use the sum of cubes identity to find:

$8 {\left(5 x + 7\right)}^{3} + 27 {\left(x - 9\right)}^{3} = 13 \left(x - 1\right) \left(79 {x}^{2} + 346 x + 1303\right)$

#### Explanation:

The sum of cubes identity may be written:

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

Use this with $a = 10 x + 14$ and $b = 3 x - 27$ as follows...

$8 {\left(5 x + 7\right)}^{3} + 27 {\left(x - 9\right)}^{3}$

$= {2}^{3} {\left(5 x + 7\right)}^{3} + {3}^{3} {\left(x - 9\right)}^{3}$

$= {\left(10 x + 14\right)}^{3} + {\left(3 x - 27\right)}^{3}$

$= \left(\left(10 x + 14\right) + \left(3 x - 27\right)\right) \left({\left(10 x + 14\right)}^{2} - \left(10 x + 14\right) \left(3 x - 27\right) + {\left(3 x - 27\right)}^{2}\right)$

$= \left(13 x - 13\right) \left(\left(100 {x}^{2} + 280 x + 196\right) - \left(30 {x}^{2} - 228 x - 378\right) + \left(9 {x}^{2} - 162 x + 729\right)\right)$

$= 13 \left(x - 1\right) \left(79 {x}^{2} + 346 x + 1303\right)$