In sawalon mein LHS se start kar ke properties apply karenge aur RHS tak pohnchenge. Aam strategy: pehle kisi row ya column ko add kar ke common factor nikalo, phir differences le kar zeros banao.
(i) \(\begin{vmatrix} 3 & 5 & 0 \\ 5 & 25 & 10 \\ 7 & 25 & 1 \end{vmatrix} = 25\begin{vmatrix} 3 & 1 & 0 \\ 1 & 1 & 2 \\ 7 & 5 & 1 \end{vmatrix}\)
Take \(5\) common from \(R_2\), then \(5\) common from \(C_2\):
\[
\text{L.H.S} = 5\begin{vmatrix} 3 & 5 & 0 \\ 1 & 5 & 2 \\ 7 & 25 & 1 \end{vmatrix} = 5 \times 5\begin{vmatrix} 3 & 1 & 0 \\ 1 & 1 & 2 \\ 7 & 5 & 1 \end{vmatrix} = 25\begin{vmatrix} 3 & 1 & 0 \\ 1 & 1 & 2 \\ 7 & 5 & 1 \end{vmatrix} = \text{R.H.S}
\]
(ii) \(\begin{vmatrix} a+x & a & a \\ a & a+x & a \\ a & a & a+x \end{vmatrix} = x^2(3a+x)\)
Apply \(R_1 \to R_1 + R_2 + R_3\) and take \((3a+x)\) common from \(R_1\):
\[
\text{L.H.S} = (3a+x)\begin{vmatrix} 1 & 1 & 1 \\ a & a+x & a \\ a & a & a+x \end{vmatrix}
\]
Apply \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\[
= (3a+x)\begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} = (3a+x) \cdot x^2 = x^2(3a+x) = \text{R.H.S}
\]
(iii) \(\begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix} = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}\)
Multiply \(R_1\) by \(x\), \(R_2\) by \(y\), \(R_3\) by \(z\) (and divide by \(xyz\)), then take \(xyz\) common from \(C_3\):
\[
\text{L.H.S} = \frac{1}{xyz}\begin{vmatrix} x & x^2 & xyz \\ y & y^2 & xyz \\ z & z^2 & xyz \end{vmatrix} = \frac{xyz}{xyz}\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}
\]
Interchange \(C_1 \leftrightarrow C_3\) and then \(C_2 \leftrightarrow C_3\) (two swaps, so no sign change):
\[
= \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = \text{R.H.S}
\]
(iv) \(\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (x-y)(y-z)(z-x)\)
Apply \(R_1 \to R_1 - R_2\) and \(R_2 \to R_2 - R_3\), then take \((x-y)\) from \(R_1\) and \((y-z)\) from \(R_2\):
\[
\text{L.H.S} = (x-y)(y-z)\begin{vmatrix} 0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & z & z^2 \end{vmatrix}
\]
Expanding along \(C_1\):
\[
= (x-y)(y-z)\begin{vmatrix} 1 & x+y \\ 1 & y+z \end{vmatrix} = (x-y)(y-z)(z-x) = \text{R.H.S}
\]
(v) \(\begin{vmatrix} 1 & 1 & 1 \\ a+1 & b+1 & c+1 \\ (a+1)^2 & (b+1)^2 & (c+1)^2 \end{vmatrix} = (a-b)(b-c)(c-a)\)
Put \(x = a+1\), \(y = b+1\), \(z = c+1\). The determinant becomes the Vandermonde form of part (iv):
\[
\text{L.H.S} = \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}
\]
Apply \(C_1 \to C_1 - C_2\) and \(C_2 \to C_2 - C_3\), take \((x-y)\) from \(C_1\) and \((y-z)\) from \(C_2\):
\[
= (x-y)(y-z)\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & z \\ x+y & y+z & z^2 \end{vmatrix}
\]
Expanding along \(R_1\) gives \((x-y)(y-z)(z-x)\). Substituting back \(x = a+1\) etc.:
\[
= (a-b)(b-c)(c-a) = \text{R.H.S}
\]
(vi) \(\begin{vmatrix} a^2+b^2 & c^2 & c^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix} = 4a^2b^2c^2\)
Apply \(R_1 \to R_1 - R_2 - R_3\) — the first row becomes \((0, -2b^2, -2a^2)\):
\[
\text{L.H.S} = \begin{vmatrix} 0 & -2b^2 & -2a^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix}
\]
Expanding along \(R_1\):
\[
= 2b^2\begin{vmatrix} a^2 & a^2 \\ b^2 & c^2+a^2 \end{vmatrix} - 2a^2\begin{vmatrix} a^2 & b^2+c^2 \\ b^2 & b^2 \end{vmatrix}
\]
\[
= 2b^2\big[a^2(c^2+a^2) - a^2b^2\big] - 2a^2\big[a^2b^2 - b^2(b^2+c^2)\big]
\]
\[
= 2a^2b^2\big[(c^2+a^2-b^2) - (a^2-b^2-c^2)\big] = 2a^2b^2(2c^2) = 4a^2b^2c^2 = \text{R.H.S}
\]
(vii) \(\begin{vmatrix} a & b & c \\ b+c & c+a & a+b \\ a+b & b+c & c+a \end{vmatrix} = a^3 + b^3 + c^3 - 3abc\)
Apply \(C_1 \to C_1 + C_2 + C_3\) and take \((a+b+c)\) common from \(C_1\):
\[
\text{L.H.S} = (a+b+c)\begin{vmatrix} 1 & b & c \\ 2 & c+a & a+b \\ 2 & b+c & c+a \end{vmatrix}
\]
Apply \(C_2 \to C_2 - bC_1\) and \(C_3 \to C_3 - cC_1\), then expand along \(R_1\):
\[
= (a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ 2 & c+a-2b & a+b-2c \\ 2 & c-b & a-c \end{vmatrix} = (a+b+c)\big[(c+a-2b)(a-c) - (a+b-2c)(c-b)\big]
\]
Expanding the bracket gives \(a^2+b^2+c^2-ab-bc-ca\). Hence:
\[
\text{L.H.S} = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc = \text{R.H.S}
\]
(viii) \(\begin{vmatrix} a+t & a & a \\ b & b+t & b \\ c & c & c+t \end{vmatrix} = t^2(a+b+c+t)\)
Apply \(R_1 \to R_1 + R_2 + R_3\) and take \((a+b+c+t)\) common from \(R_1\):
\[
\text{L.H.S} = (a+b+c+t)\begin{vmatrix} 1 & 1 & 1 \\ b & b+t & b \\ c & c & c+t \end{vmatrix}
\]
Apply \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\), then expand along \(R_1\):
\[
= (a+b+c+t)\begin{vmatrix} 1 & 0 & 0 \\ b & t & 0 \\ c & 0 & t \end{vmatrix} = (a+b+c+t)(t^2) = t^2(a+b+c+t) = \text{R.H.S}
\]
(ix) \(\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^3\)
Apply \(R_1 \to R_1 + R_2 + R_3\) and take \((a+b+c)\) common from \(R_1\):
\[
\text{L.H.S} = (a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}
\]
Apply \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\):
\[
= (a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix}
\]
Expanding along \(R_1\):
\[
= (a+b+c)\big[-(a+b+c)\big]\big[-(a+b+c)\big] = (a+b+c)^3 = \text{R.H.S}
\]
(x) \(\begin{vmatrix} y+z & z+x & x+y \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} = (x+y+z)(x-y)(y-z)(z-x)\)
Apply \(R_1 \to R_1 + R_2\) — the first row becomes \((x+y+z, x+y+z, x+y+z)\). Take \((x+y+z)\) common:
\[
\text{L.H.S} = (x+y+z)\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}
\]
This is the Vandermonde determinant of part (iv) (with variables in rows), which equals \((x-y)(y-z)(z-x)\). Hence:
\[
\text{L.H.S} = (x+y+z)(x-y)(y-z)(z-x) = \text{R.H.S}
\]
(xi) \(\begin{vmatrix} 1 & 1 & 1 \\ a^2+1 & b^2+1 & c^2+1 \\ a^3+a & b^3+b & c^3+c \end{vmatrix} = (a-b)(b-c)(c-a)(ab+bc+ca-1)\)
Apply \(C_1 \to C_1 - C_2\) and \(C_2 \to C_2 - C_3\). Using \(a^2 - b^2 = (a-b)(a+b)\) and \(a^3 + a - b^3 - b = (a-b)(a^2+ab+b^2+1)\):
\[
\text{L.H.S} = \begin{vmatrix} 0 & 0 & 1 \\ a^2-b^2 & b^2-c^2 & c^2+1 \\ (a-b)(a^2+ab+b^2+1) & (b-c)(b^2+bc+c^2+1) & c^3+c \end{vmatrix}
\]
Take \((a-b)\) common from \(C_1\) and \((b-c)\) common from \(C_2\), then expand along \(R_1\):
\[
= (a-b)(b-c)\begin{vmatrix} a+b & b+c \\ a^2+ab+b^2+1 & b^2+bc+c^2+1 \end{vmatrix}
\]
\[
= (a-b)(b-c)\big[(a+b)(b^2+bc+c^2+1) - (b+c)(a^2+ab+b^2+1)\big]
\]
Expanding and simplifying the bracket gives \((c-a)(ab+bc+ca-1)\). Therefore:
\[
\text{L.H.S} = (a-b)(b-c)(c-a)(ab+bc+ca-1) = \text{R.H.S}
\]
(xii) \(\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc + ab + bc + ca\)
Apply \(R_1 \to R_1 - R_2\) and \(R_2 \to R_2 - R_3\):
\[
\text{L.H.S} = \begin{vmatrix} a & -b & 0 \\ 0 & b & -c \\ 1 & 1 & 1+c \end{vmatrix}
\]
Expanding along \(R_1\):
\[
= a\big[b(1+c) + c\big] + b\big[0 + c\big] = a(b + bc + c) + bc
\]
\[
= ab + abc + ac + bc = abc + ab + bc + ca = \text{R.H.S}
\]
(xiii) \(\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \end{vmatrix} = 0\)
Apply \(R_1 \to R_1 - R_2\) and \(R_2 \to R_2 - R_3\). Since \(a^2 - b^2 - bc + ca = (a-b)(a+b+c)\):
\[
\text{L.H.S} = \begin{vmatrix} 0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & c^2-ab \end{vmatrix}
\]
Take \((a-b)\) common from \(R_1\) and \((b-c)\) common from \(R_2\):
\[
= (a-b)(b-c)\begin{vmatrix} 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^2-ab \end{vmatrix}
\]
Since \(R_1\) and \(R_2\) are identical, the determinant is \(0\).
(xiv) \(\begin{vmatrix} 2x+3 & x+2 & x+a \\ 2x+5 & x+3 & x+b \\ 2x+7 & x+4 & x+c \end{vmatrix} = 0\), where \(2b = a+c\)
Apply \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_2\):
\[
\text{L.H.S} = \begin{vmatrix} 2x+3 & x+2 & x+a \\ 2 & 1 & b-a \\ 2 & 1 & c-b \end{vmatrix}
\]
Apply \(R_3 \to R_3 - R_2\):
\[
= \begin{vmatrix} 2x+3 & x+2 & x+a \\ 2 & 1 & b-a \\ 0 & 0 & c-2b+a \end{vmatrix}
\]
Since \(2b = a + c\), we have \(c - 2b + a = 0\), so \(R_3\) becomes a zero row. Hence the determinant is \(0\).
(xv) \(\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)\)
Apply \(C_1 \to C_1 + C_2 + C_3\) and take \((a+b+c)\) common from \(C_1\):
\[
\text{L.H.S} = (a+b+c)\begin{vmatrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{vmatrix}
\]
Apply \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\), then expand along \(C_1\):
\[
= (a+b+c)\begin{vmatrix} 1 & b & c \\ 0 & a-b & b-c \\ 0 & c-b & a-c \end{vmatrix} = (a+b+c)\big[(a-b)(a-c) - (b-c)(c-b)\big]
\]
\[
= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
\]
Using the cube roots of unity (\(\omega^3 = 1\), \(1 + \omega + \omega^2 = 0\)):
\[
a^2+b^2+c^2-ab-bc-ca = (a+b\omega+c\omega^2)(a+b\omega^2+c\omega)
\]
Therefore \(\text{L.H.S} = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) = \text{R.H.S}\).
Answer: All fifteen identities are proved using determinant properties.
