Question

The value of a for which the system of equations $a^{3}x+(a+1{)}^{3}y+(a+2{)}^{3}z=0$, $ax+(a+1)y+(a+2)z=0$, $x+y+z=0$ has a non zero solution is?

Answer 1

The system has non zero solution if,

$\left|\begin{array}{ccc}{a}^3& (a+1{)}^3& (a+2{)}^3\\ a& a+1& a+2\\ 1& 1& 1\end{array}\right|=0$

$C_2\to C_2-C_1,C_3\to C_3-C_2$

$\left|\begin{array}{ccc}{a}^3& (a+1{)}^3-a^3& (a+2{)}^3-(a+1{)}^3\\ a& 1& 1\\ 1& 0& 0\end{array}\right|$

$a^3\left(0\right)-\left[\right(a+1{)}^3-a^3\left]\right(0-1)+[(a+2{)}^3-(a+1{)}^3\left]\right(0-1)=0$

$(a+1{)}^3-a^3-(a+2{)}^3+(a+1{)}^3=0$

$2(a+1{)}^3-a^3-(a+2{)}^3=0$

$2(a^3+3a^2+3a+1)-a^3-(a^3+6a^2+12a+8)=0$

$2a^3+6a^2+6a+2-a^3-a^3-6a^2-12a-8=0$

$2a^3-a^3-a^3-6a+2-8=0$

$-6a+2-8=0$

$-6a-6=0$

$\therefore a=-1$