Question

[tex] \left |\begin{array}{ccc} x&y&z \\x-y&y-z&z-x \\ y+z& z+x &x+y\end{array}\right | =x³+y³+z³-3xyz,prove it using theorems [\tex]

Answer 1

Step-by-step explanation:

$\left |\begin{array}{ccc} x&y&z \\x-y&y-z&z-x \\ y+z& z+x &x+y\end{array}\right | =x^3+y^3+z^3-3xyz$

To prove it using theorems

Apply

$c1 = c1 + c2 + c3$

$\left |\begin{array}{ccc} x+y+z&y&z \\x-y+y-z+z-x&y-z&z-x \\ y+z+z+x+x+y& z+x &x+y\end{array}\right |$

$\left |\begin{array}{ccc} x+y+z&y&z \\0&y-z&z-x \\2(x+y+z)& z+x &x+y\end{array}\right |$

take common (x+y+z) from C1

$(x+y+z)\left |\begin{array}{ccc} 1&y&z \\0&y-z&z-x \\2& z+x &x+y\end{array}\right |$

R3->R3-2R1

$(x+y+z)\left |\begin{array}{ccc} 1&y&z \\0&y-z&z-x \\0& z+x-2y &x+y-2z\end{array}\right |$

Now expand the determinant along Column 1

$(x + y + z)((y - z)(x + y - 2z) - (z - x)(z + x - 2y)) \\ \\ (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - xz) \\ \\ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

= RHS

Hope it helps you.