If

$y=c_1{e}^{2x}+c_2{e}^x+c_3{e}^{-x}$

satisfies the differential equation

$\frac{d^{3}y}{d{x}^3}+a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0$

then

$\frac{a^3+b^3+c^3}{abc}=$

a$)1$

b$)-\frac{1}{4}$

c$)-\frac{1}{2}$

d$)0$

e$)\frac{1}{2}$

Bos birak