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$A=\begin{bmatrix} a&a^{2} &a^{3}-1 \\ b& b^{2} &b^{3}-1 \\ c& c^{2}&c^{3}-1 \end{bmatrix}$

$Given |A|=\begin{vmatrix} a&a^{2} &a^{3}-1 \\ b& b^{2} &b^{3}-1 \\ c& c^{2}&c^{3}-1 \end{vmatrix}=0$

$\begin{vmatrix} a&a^{2} &a^{3} \\ b& b^{2} &b^{3} \\ c& c^{2}&c^{3} \end{vmatrix}+\begin{vmatrix} a&a^{2} &-1 \\ b& b^{2} &-1 \\ c& c^{2}&-1 \end{vmatrix}=0$

$abc\begin{vmatrix} 1&a &a^{2} \\ 1& b&b^{2} \\ 1& c&c^{2} \end{vmatrix}+\begin{vmatrix} a&a^{2} &-1 \\ b& b^{2} &-1 \\ c& c^{2}&-1 \end{vmatrix}=0$

$abc\begin{vmatrix} 1&a &a^{2} \\ 1& b&b^{2} \\ 1& c&c^{2} \end{vmatrix}+\begin{vmatrix} a&-1&a^{2} \\ b& -1&b^{2} \\ c& -1&c^{2} \end{vmatrix}=0$

$abc\begin{vmatrix} 1&a &a^{2} \\ 1& b&b^{2} \\ 1& c&c^{2} \end{vmatrix}+\begin{vmatrix} -1&a&a^{2} \\ -1& b&b^{2} \\ -1 & c&c^{2} \end{vmatrix}=0$

$abc\begin{vmatrix} 1&a &a^{2} \\ 1& b&b^{2} \\ 1& c&c^{2} \end{vmatrix}-1\begin{vmatrix} 1&a&a^{2} \\ 1& b&b^{2} \\ 1 & c&c^{2} \end{vmatrix}=0$

$\begin{vmatrix} 1&a &a^{2} \\ 1& b&b^{2} \\ 1& c&c^{2} \end{vmatrix}(abc-1)=0$

$\begin{vmatrix} 1&a &a^{2} \\ 1& b&b^{2} \\ 1& c&c^{2} \end{vmatrix}=0 , (abc-1)=0$

$abc=1$
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1 votes
1 votes
Vale of ABC is 1

Solution:::::

Divide the Matrix at third column

Take a from first row , b from second row , c from third row

Swap first determinant first column and second then second and third

Then take common det and u will be left with abc-1 ::0

Hence ABC :: 1

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