Factoring a sum or difference of cubes:
(x+5)^3-(2x+1)^3

Use difference of cubes: a³-b³=(a-b)(a²+ab+b²)
(4-x)((x+5)²+(x+5)(2x+1)+(2x+1)²)

(x + 5)³ - (2x + 1)³

Using the formula a³ - b³ = (a - b) (a² + ab + b²) , we get,

((x+5) - (2x + 1)) ((x+5)² + (x+5)(2x + 1) + (2x + 1)²)

Using formula of (a + b)² = a² + 2ab + b² , we get,

= (x + 5 - 2x - 1) (x² + 10x + 25 + 2x² + x + 10x + 5 + 4x² + 4x + 1)
= (-x +4) (7x² + 25x + 31)
= -7x³ - 25x² - 31x + 28x² + 100x + 124
= -7x³ + 3x² + 69x + 124  > thats the answer