If a, b, c are unequal,what is the condition | vedclass

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(A) We are given the determinant $\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 + 1 \ b & b^2 & b^3 + 1 \ c & c^2 & c^3 + 1 \end{array} ight|$.

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$1 + abc = 0$

Vedclass · Answer

(A) We are given the determinant $\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 + 1 \ b & b^2 & b^3 + 1 \ c & c^2 & c^3 + 1 \end{array} ight|$.By the property of determinants,we can split the third column into two parts:$\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3 \end{array} ight| + \left| \begin{array}{ccc} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array} ight|$.Taking $abc$ common from the first determinant,we get:$\Delta = abc \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| + \left| \begin{array}{ccc} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array} ight|$.In the second determinant,perform column swaps to match the first: $C_3 \leftrightarrow C_2$ then $C_2 \leftrightarrow C_1$. This results in two sign changes,so the sign remains positive:$\Delta = abc \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| + \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| = (1 + abc) \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight|$.The value of the determinant $\left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight|$ is $(a - b)(b - c)(c - a)$.Thus,$\Delta = (1 + abc)(a - b)(b - c)(c - a) = 0$.Since $a, b, c$ are unequal,$(a - b)(b - c)(c - a) 
eq 0$.Therefore,the condition is $1 + abc = 0$.

If $a, b, c$ are unequal,what is the condition that the value of the following determinant is zero? $\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 + 1 \\ b & b^2 & b^3 + 1 \\ c & c^2 & c^3 + 1 \end{array} \right|$

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