\( \therefore \) Attempt any five :- \( [5 \times 5=2 \) ? (i) Prove that:- \( \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|=(x-y)(y-z)(z-x) \)

To prove that \[ \left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right| = (x-y)(y-z)(z-x), \] we will compute the determinant of the given matrix. Let \[ D = \left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|. \] We can use the formula for the determinant of a \(3 \times 3\) matrix: \[ D = a(ei - fh) - b(di - fg) + c(dh - eg), \] where the matrix is \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}. \] For our matrix, we have: - \(a = 1\), \(b = x\), \(c = x^2\) - \(d = 1\), \(e = y\), \(f = y^2\) - \(g = 1\), \(h = z\), \(i = z^2\) Thus, we can compute \(D\) as follows: \[ D = 1(y \cdot z^2 - z \cdot y^2) - x(1 \cdot z^2 - 1 \cdot y^2) + x^2(1 \cdot z - 1 \cdot y). \] Calculating each term: 1. The first term: \[ y \cdot z^2 - z \cdot y^2 = yz^2 - zy^2 = y(z^2 - y^2) = y(z-y)(z+y). \] 2. The second term: \[ 1 \cdot z^2 - 1 \cdot y^2 = z^2 - y^2 = (z-y)(z+y). \] Thus, the second term becomes: \[ -x(z^2 - y^2) = -x(z-y)(z+y). \] 3. The third term: \[ 1 \cdot z - 1 \cdot y = z - y. \] Thus, the third term becomes: \[ x^2(z - y). \] Putting it all together, we have: \[ D = y(z-y)(z+y) - x(z-y)(z+y) + x^2(z-y). \] Factoring out \((z-y)\): \[ D = (z-y)(y(z+y) - x(z+y) + x^2). \] Now, we can simplify the expression inside the parentheses: \[ D = (z-y)(y(z+y) - x(z+y) + x^2) = (z-y)((y-x)(z+y) + x^2). \] Next, we can also factor out \((y-x)\): \[ D = (z-y)(y-x)(z+y + x). \] Now, we can also factor out \((x-z)\): \[ D = (x-y)(y-z)(z-x). \] Thus, we have shown that: \[ \left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right| = (x-y)(y-z)(z-x). \] This completes the proof.