Suppose $a, b, c$ are three integers such $a < b < c$ and $p$is a prime number.

Let       $\mathrm{\Delta }=\left|\begin{array}{ccc}a& a^2& p+a^3\\ b& b^2& p+b^3\\ c& c^2& p+c^3\end{array}\right|$

If $∆ = 0,$ then which one of the following is not true.

Write $\mathrm{\Delta }=p{\mathrm{\Delta }}_1+{\mathrm{\Delta }}_2$

Where

${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|$and ${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|$

write

${\mathrm{\Delta }}_2=abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=abc(-1{)}^2\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|$

                     $=abc{\mathrm{\Delta }}_1$

Thus      $\mathrm{\Delta }=(p+abc){\mathrm{\Delta }}_1$

Using $R_2\to R_2-R_3,R_1\to R_1-R_2$ we get

$\begin{array}{r}{\mathrm{\Delta }}_1=\left|\begin{array}{ccc}a-b& a^2-b^2& 0\\ b-c& b^2-c^2& 0\\ c& c^2& 1\end{array}\right|\\ =(a-b)(b-c)\left|\begin{array}{cc}1& a+b\\ 1& b+c\end{array}\right|\\ =(a-b)(b-c)(c-a)\end{array}$

As $a<b<c,{\mathrm{\Delta }}_1\ne 0$ Therefore,

             $\mathrm{\Delta }=0\Rightarrow p+abc=0$

$\Rightarrow p=-abc$

As $p$ is prime and $a, b, c$ are integers such that $a < b < c,$ we must have $-a=1,b=1,c=p.$

$\Rightarrow a=-1,b=1,c=p$