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If the system of equations $(k + 1) x + {(k + 2)}^{3} y = {(k + 3)}^{3}$ , $(k + 1) x + (k + 2) y = k + 3$ , $x + y = 1$ is consistent, then k=__________

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-2

-1

answer is B.

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Detailed Solution

$\Rightarrow |\begin{matrix} a^{3} & b^{3} & c^{3} \\ a & b & c \\ 1 & 1 & 1 \end{matrix}| = 0$ where $a = k + 1, b = k + 2, c = k + 3$

$\Rightarrow (a - b) (b - c) (c - a) [a + b + c] = 0$

$\Rightarrow$ $a + b + c = 0 \Rightarrow k + 1 + k + 2 + k + 3 = 0$

$\Rightarrow$ $3 k = - 6 \Rightarrow k = - 2$

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