Question

if x y z are different and
$\mathtt \green{\Delta=\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0}$
then prove that 1+xyz=0.
Answer with proper explanation.

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x \: \ne \: y \: \ne \: z$

and

$\rm :\longmapsto\:\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0$

We know,

In determinants, if any column is represented as sum of two or more elements, then determinant can also be expressed as sum of two or more determinants.

So, above determinant can also be rewritten as

$\rm :\longmapsto\:\left|\begin{array}{ccc}x & x^{2} & 1\\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| + \left|\begin{array}{ccc}x & x^{2} &x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|=0$

From Second determinant, take out x, y and z common from respective Rows.

$\rm :\longmapsto\:\left|\begin{array}{ccc}x & x^{2} & 1\\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| + xyz\left|\begin{array}{ccc}1 & x^{} &x^{2} \\ 1 & y^{} & y^{2} \\ 1 & z^{} & z^{2}\end{array}\right|=0$

In first determinant,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: C_2 \: \leftrightarrow \: C_3}}$

$\rm :\longmapsto\: - \left|\begin{array}{ccc}x & 1 & {x}^{2} \\ y & 1 & {y}^{2} \\ z & 1 & {z}^{2} \end{array}\right| + xyz\left|\begin{array}{ccc}1 & x^{} &x^{2} \\ 1 & y^{} & y^{2} \\ 1 & z^{} & z^{2}\end{array}\right|=0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: C_2 \: \leftrightarrow \: C_1}}$

$\rm :\longmapsto\:\left|\begin{array}{ccc}1 & x & {x}^{2} \\ 1 & y & {y}^{2} \\ 1 & z & {z}^{2} \end{array}\right| + xyz\left|\begin{array}{ccc}1 & x^{} &x^{2} \\ 1 & y^{} & y^{2} \\ 1 & z^{} & z^{2}\end{array}\right|=0$

can be rewritten as after taking common,

$\rm :\longmapsto\:(1 + xyz)\left|\begin{array}{ccc}1 & x & {x}^{2} \\ 1 & y & {y}^{2} \\ 1 & z & {z}^{2} \end{array}\right| =0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: R_2 \: \rightarrow \:R_2 - R_1}}$

and

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: R_3 \: \rightarrow \:R_3 - R_1}}$

$\rm :\longmapsto\:(1 + xyz)\left|\begin{array}{ccc}1 & x & {x}^{2} \\ 0 & y - x& {y}^{2} - {x}^{2} \\ 0 & z - x & {z}^{2} - {x}^{2} \end{array}\right| =0$

$\rm :\longmapsto\:(1 + xyz)\left|\begin{array}{ccc}1 & x & {x}^{2} \\ 0 & y - x & (y - x)(y + x) \\ 0 & z - x & (z - x)(z + x) \end{array}\right| =0$

$\rm :\longmapsto\:(1 + xyz)(y - x)(z - x)\left|\begin{array}{ccc}1 & x & {x}^{2} \\ 0 &1 & y + x \\ 0 & 1 & z + x \end{array}\right| =0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: R_3 \: \rightarrow \:R_3 - R_2}}$

$\rm :\longmapsto\:(1 + xyz)(y - x)(z - x)\left|\begin{array}{ccc}1 & x & {x}^{2} \\ 0 &1 & y + x \\ 0 & 0 & z - y \end{array}\right| =0$

$\rm :\longmapsto\:(1 + xyz)(y - x)(z - x)(z - y)=0$

As it is given that,

$\rm :\longmapsto\:x \: \ne \: y \: \ne \: z$

THUS,

$\bf\implies \:1 + xyz = 0$

Hence, Proved

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More to know :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.