Sum of Cubes

Lets start with $8x^3 + 27 = (2x)^3 + 3^3$ $= (2x + 3)((2x)^2 - (2*3x) + 3^2)$ $= (2x+3)(4x^2 - 6x + 9)$ Summing the coefficents we get $24$

let $8x^{3}$ + 27 = 0, so ${x}$ = - $\frac{3}{2}$ . This rearranges to give 2 ${x}$ + 3 = 0 = ${a}$ ${x}$ + ${b}$ . By inspection $8x^{3}$ + 27 = (2 ${x}$ +3)(4 $x^{2}$ - d ${x}$ +9). By comparing coefficients of ${x}$ : 18 - 3 ${d}$ = 0, so ${d}$ =6. This means that the required factorization is: (2 ${x}$ +3)(4 $x^{2}$ - 6 ${x}$ +9). $\therefore$ ${a}$ + ${b}$ + ${c}$ + ${d}$ + ${e}$ = 24

use the formula = a^{3} + b^{3} = (a + b) \times (a^{2} + b ^{2} - ab )