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The value of ${ }^{\prime} a$ for which the system of equations
$$
\begin{array}{r}
a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0 \\
a x+(a+1) y+(a+2) z=0 \\
x+y+z=0
\end{array}
$$
has a non-zero solution is
Options:
$$
\begin{array}{r}
a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0 \\
a x+(a+1) y+(a+2) z=0 \\
x+y+z=0
\end{array}
$$
has a non-zero solution is
Solution:
2182 Upvotes
Verified Answer
The correct answer is:
$-1$
For non-zero solution,
$$
\left|\begin{array}{ccc}
a^{3} & (a+1)^{3} & (a+2)^{3} \\
a & (a+1) & (a+2) \\
1 & 1 & 1
\end{array}\right|=0
$$
$$
\left.\begin{array}{l}
\Rightarrow \quad-\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & (a+1) & (a+2) \\
a^{3} & (a+1)^{3} & (a+2)^{3}
\end{array}\right|=0 \\
\Rightarrow & -(a-a-1)(a+1-a-2)(a+2-a) \\
& \quad \quad \times(a+a+1+a+2)=0 \\
\Rightarrow & -2(3 a+3)=0 \\
\Rightarrow & \quad a=-1 \\
& {\left[\begin{array}{ccc}
1 & 1 & 1 \\
x & y & z \\
x^{3} & y^{3} & z^{3}
\end{array} \mid=\right.} \\
(x-y)(y-z)(z-x)(x+y+z)
\end{array}\right]
$$
$$
\left|\begin{array}{ccc}
a^{3} & (a+1)^{3} & (a+2)^{3} \\
a & (a+1) & (a+2) \\
1 & 1 & 1
\end{array}\right|=0
$$
$$
\left.\begin{array}{l}
\Rightarrow \quad-\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & (a+1) & (a+2) \\
a^{3} & (a+1)^{3} & (a+2)^{3}
\end{array}\right|=0 \\
\Rightarrow & -(a-a-1)(a+1-a-2)(a+2-a) \\
& \quad \quad \times(a+a+1+a+2)=0 \\
\Rightarrow & -2(3 a+3)=0 \\
\Rightarrow & \quad a=-1 \\
& {\left[\begin{array}{ccc}
1 & 1 & 1 \\
x & y & z \\
x^{3} & y^{3} & z^{3}
\end{array} \mid=\right.} \\
(x-y)(y-z)(z-x)(x+y+z)
\end{array}\right]
$$
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