(B) False. Let $\alpha, \beta$ and $\gamma$ be the three zeroes of the cubic polynomial $p(x) = x^{3}+ax^{2}-bx+c$.
According to the relationship between zeroes and coefficients:
$1$. Product of zeroes: $\alpha \beta \gamma = -\frac{c}{1} = -c$. Since $\alpha, \beta, \gamma > 0$,their product $\alpha \beta \gamma > 0$. Therefore,$-c > 0$,which implies $c < 0$.
$2$. Sum of zeroes: $\alpha + \beta + \gamma = -\frac{a}{1} = -a$. Since $\alpha, \beta, \gamma > 0$,their sum $\alpha + \beta + \gamma > 0$. Therefore,$-a > 0$,which implies $a < 0$.
$3$. Sum of product of zeroes taken two at a time: $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{-b}{1} = -b$. Since $\alpha, \beta, \gamma > 0$,their sum of products $\alpha \beta + \beta \gamma + \gamma \alpha > 0$. Therefore,$-b > 0$,which implies $b < 0$.
Thus,for all three zeroes to be positive,$a, b$ and $c$ must all be negative. Hence,the statement that at least one of them is non-negative is False.