We start from the factorization by the roots. If the cubic
\[
ax^3+bx^2+cx+d
\]
has roots \(r_1,r_2,r_3\), then (up to the leading coefficient)
\[
ax^3+bx^2+cx+d = a(x-r_1)(x-r_2)(x-r_3).
\]
Expand the right-hand side:
\[
(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3.
\]
Multiplying by \(a\) gives
\[
ax^3 - a(r_1+r_2+r_3)x^2 + a(r_1r_2+r_1r_3+r_2r_3)x - a\,r_1r_2r_3.
\]
Compare coefficients with \(ax^3+bx^2+cx+d\):
- Coefficient of \(x^2\): \(\;b = -a(r_1+r_2+r_3)\;\) so \(\;r_1+r_2+r_3 = -\dfrac{b}{a}.\)
- Coefficient of \(x\): \(\;c = a(r_1r_2+r_1r_3+r_2r_3)\;\) so \(\;r_1r_2+r_1r_3+r_2r_3 = \dfrac{c}{a}.\)
- Constant term: \(\;d = -a\,r_1r_2r_3\;\) so \(\;r_1r_2r_3 = -\dfrac{d}{a}.\)

Therefore the correct Vieta relations are
\[
\boxed{\,r_1+r_2+r_3 = -\dfrac{b}{a},\qquad r_1r_2+r_1r_3+r_2r_3 = \dfrac{c}{a},\qquad r_1r_2r_3 = -\dfrac{d}{a}\,}
\]
This corresponds to Option 2. The alternative list you gave (with the pairwise sum \(-c/a\) and product \(d/a\)) has the signs reversed and is not correct for the polynomial written as \(ax^3+bx^2+cx+d\).