Question
If the equations $(a+1)^{3} x+(a+2)^{3} y=(a+3)^{3},(a+1) x+(a+2) y$$=a+3, x+y=1$ are consistent then $a$ is equal to(A) 1(B) $-1$(C) 2(D) $-2$
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$y=x^{3}+a x$ $\frac{d y}{d x}=3 x^{2}+a$ $\frac{d y}{d x}=3+a$ $y=b x^{2}+c$ $\frac{d y}{d x}=2 b x=-2 b$ $\Rightarrow 3+a=-2 b$ $\Rightarrow 3+a+2 b=0$ As $-1-a=0$ $\Rightarrow a=-1$ $b+c=0$ $b=-1$ $c=1$ Then, $\left(a+b+c^{2}\right)=-1-1+1=-1$
$$\begin{array}{l}{y=x^{2}+1} \\ {y=-x^{2}+3}\end{array}$$ If $(a, b)$ and $(c, d)$ are solutions of the system of equations above, what is the value of $a+b+c+d ?$ $$\begin{array}{l}{\text { (A) } 1} \\ {\text { (B) } 2} \\ {\text { (C) } 3} \\ {\text { (D) } 4}\end{array}$$
If no two of $a, b, c$ are equal and $a x+a^{2} y+a^{3}+1=0, b x+b^{2} y+b^{3}+1=0, c x+c^{2} y+c^{3}+1=0$, then (a) $a b c=1$ (b) $a b c=-1$ (c) $\mathrm{abc}=0$ (d) $a=1, b=2, c=3$
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