If $(x+6)^3+(2 x+3)^3+(3 x+5)^3=(3 x+18)(2 x+3)(3 x+5)$, then what is the value of $x$ ?

$-\frac{5}{3}$ $\frac{5}{3}$ $-\frac{7}{3}$ $\frac{7}{3}$

$-\frac{7}{3}$

(x + 6)3 + (2x + 3)3 + (3x + 5)3 = (3x + 18)(2x + 3)(3x + 5) = (x + 6)3 + (2x + 3)3 + (3x + 5)3 = 3(x + 6) (2x + 3) (3x + 5) Now, If (a + b + c) = 0 then,  a3 + b3 + c3 = 3abc = (x + 6) + (2x + 3) + (3x + 5) = 0 = 6x + 14 = 0 = 6x = -14 x = $-\frac{7}{3}$