Question

a p Xa +b p+q x+y

Answer 1

The left hand side determinant is,

Δ=| b+c q+r y+z c+a r+p z+x a+b p+q x+y |

Apply row operation, R 3 → R 1 +R 2 + R 3 .

Δ=| b+c q+r y+z c+a r+p z+x a+b+b+c+c+a p+q+q+r+r+p x+y+y+z+z+x | =| b+c q+r y+z c+a r+p z+x 2( a+b+c ) 2( p+q+r ) 2( x+y+z ) |

Apply row operation, R 1 → R 1 − R 3 .

Δ=2| b+c−a−b−c q+r−p−q−r y+z−x−y−z c+a r+p z+x ( a+b+c ) ( p+q+r ) ( x+y+z ) | =2| −a −p −x c+a r+p z+x ( a+b+c ) ( p+q+r ) ( x+y+z ) |

Apply row operation, R 2 → R 2 − R 3 .

Δ=2| −a −p −x c+a−a−b−c r+p−p−q−r z+x−x−y−z ( a+b+c ) ( p+q+r ) ( x+y+z ) | =2| −a −p −x −b −q −y ( a+b+c ) ( p+q+r ) ( x+y+z ) |

Apply row operation, R 3 → R 3 +R 1 + R 2 .

Δ=2| −a −p −x −b −q −y a+b+c−a−b p+q+r−p−q x+y+z−x−y | =2| −a −p −x −b −q −y c r z | =2| a p x b q y c r z |

From above determinants, left hand side and right hand side are equal.

Hence, it is proved.