Chapter 8, Problem 11PS

To determine

Verify the equation.

Answer to Problem 11PS

It is proved that $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Explanation of Solution

Given:

The equation

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

Considertheequation

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

Substitute $a=b$ .Therefore,thedeterminantbecomes,

  $\left|\begin{array}{ccc}1& 1& 1\\ b& b& c\\ b^3& b^3& c^3\end{array}\right|$

Sincetwocolumnsofthedeterminantaresame,thedeterminanthasvaluezero.

Therefore,oneofthefactorsofthedeterminantis $a–b$ .

Substitute $b=c$ . Therefore, the determinant becomes,

  $\left|\begin{array}{ccc}1& 1& 1\\ a& c& c\\ a^3& c^3& c^3\end{array}\right|$

Since two columns of the determinant are same, the determinant has value zero.

Therefore, one of the factors of the determinant is $b–c$ .

Substitute $c=a$ . Therefore, the determinant becomes,

  $\left|\begin{array}{ccc}1& 1& 1\\ a& b& a\\ a^3& b^3& a^3\end{array}\right|$

Since two columns of the determinant are same, the determinant has value zero.

Therefore, one of the factors of the determinant is $c–a$ .

The polynomial in the determinant is of degree 4. Therefore,

  $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=k(a-b)(b-c)(c-a)(a+b+c)$

Substitute $a=2,b=1,c=0$ . Therefore, equation becomes

  $\begin{array}{l}\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 0\\ 1& 8& 0\end{array}\right|=k(1-2)(2-0)(0-1)(1+2+0)\\ \begin{array}{l}1\left(8–2\right)=k\left(–1\right)\left(2\right)\left(–1\right)\left(3\right)\hfill \\ 6=6k\hfill \\ k=1\hfill \end{array}\end{array}$

Therefore, it is proved that $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$