Question

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

• A. True
• B. False

Answer 1

The correct option is B False

$\left|\begin{array}{ccc}1& a& a^3\\ 1& b& b^3\\ 1& c& c^3\end{array}\right|=\left|\begin{array}{ccc}1& a& a^3\\ 0& b-a& b^3-a^3\\ 0& c-a& c^3-a^3\end{array}\right|\{\begin{array}{c}{R}_2^{{}^{\prime }}=R_2-R_1\\ R_3^{{}^{\prime }}=R_3-R_1\end{array}=\left|\begin{array}{ccc}1& a& a^3\\ 0& b-a& (b-a)(a^2+ba+b^2)\\ 0& c-a& (c-a)(c^2+ac+a^2)\end{array}\right|\{\begin{array}{c}\text{Formula of}{a}^3-b^3\end{array}=(b-a)(c-a)\left|\begin{array}{ccc}1& a& a^3\\ 0& b-a& (a^2+ba+b^2)\\ 0& (c-a)& (c^2+ac+a^2)\end{array}\right|=(b-a)(c-a)(c^2+ac+a^2-a^2-ab-b^2)\text{[expanding with w.r.t. to Ist row]}(b-a)(c-a)(c^2-b^2+a(c-b\left)\right)=-(a-b)(c-a)(c-b)(a+b+c)$

Hence the given answer is wrong (false)