Question

For what value of p, is the system of equations :
$p^{3}x+(p+1{)}^{3}y=(p+2{)}^3$
$px+(p+1)y=p+2$
$x+y=1$
consistent ?

• A. p = 0
• B. p = 1
• C. p = - 1
• D. For all p > 1

Answer 1

Answer: (C)

The given system of equations are :
$p^{3}x+{\left(p+1\right)}^{3}y={\left(p+2\right)}^3...\left(1\right)$
$px+\left(p+1\right)y=p+2...\left(2\right)$
$x+y=1...\left(3\right)$
This system is consistent, if values of x and y from first two equation satisfy the third equation.
which $\Rightarrow \left|\begin{array}{ccc}{p}^3& {\left(p+1\right)}^3& {\left(p+2\right)}^3\\ p& \left(p+1\right)& \left(p+2\right)\\ 1& 1& 1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}{p}^3& {\left(p+1\right)}^3-p^3& {\left(p+2\right)}^3-p^3\\ p& 1& 2\\ 1& 0& 0\end{array}\right|=0$
$\Rightarrow 2{\left(p+1\right)}^3-2p^3-{\left(p+2\right)}^3+p^3=0$
$\Rightarrow 2\left(p^3+1+3p^2+3p\right)-2p^3-\left(p^3+8+12p+6p^2\right)+p^3=0$
$\Rightarrow 2p^3+2+6p^2+6p-2p^3-p^3-8-12p-6p^2+p^3=0$
$\Rightarrow -6-6p=0$
$\Rightarrow p=-1$