Vieta's Theorems Clarification

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For applying Vieta's Theorems on cubic equations, I've seen two conflicting answers from two different books.

Book 1 says that for a cubic equation $px^3 + qx^2 + rx + s = 0$ has three different roots $x_1, x_2, x_3$, the following is true: $$\begin{eqnarray*} -q &=& x_1 + x_2 + x_3 \\ \end{eqnarray*}$$ Book 2, however, says that for a cubic equation $px^3 + qx^2 + rx + s = 0$ has three different roots $x_1, x_2, x_3$, this is true: $$\begin{eqnarray*} x_1 + x_2 + x_3 &=& \frac qp \\ \end{eqnarray*}$$

Can anyone tell me which one is correct?

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Neither is correct. It should be $x_1+x_2+x_3=-q/p$.

The cubic has shape $p(x-x_1)(x-x_2)(x-x_3)$. Expanding we get $$px^3-p(x_1+x_2+x_3)x^2+p(x_1x_2+x_2x_3+x_3x_1)x-p(x_1x_2x_3).$$ Comparing coefficients, we get $$x_1+x_2+x_3=-q/p,\quad x_1x_2+x_2x_3+x_3x_1=r/p, \quad x_1x_2x_3=-s/p.$$