सिद्ध कीजिए कि `|{:((x+y)^(2),zx,zy),(zx,(z+y)^(2),xy),(zy,xy,(z+x)^(2)):}|=2xyz(x+y+z)^(3)`

Explanation

बायाँ पक्ष `|{:((x+y)^(2),zx,zy),(zx,(z+y)^(2),xy),(zy,xy,(z+x)^(2)):}|`
`=(1)/(xyz)|{:(z(x+y)^(2),z^(2)x,z^(2)y),(zx^(2),x(z+y)^(2),x^(2)y),(zy^(2),xy^(2),y(z+x)^(2)):}| (R_(1)tozR_(1),R_(2)toxR_(2)` तथा `R_(3)toyR_(3))`
`=(xyz)/(xyz)|{:((x+y)^(2),z^(2),z^(2)),(x^(2),(z+y)^(2),x^(2)),(y^(2),y^(2),(z+x)^(2)):}|C_(1)to(1)/(z)C_(1),C_(2)to(1)/(x)C_(2)` तथा `C_(3)to(1)/(y)C_(3)`
`=|{:((x+y)^(2)-z^(2),0,z^(2)),(0,(z+y)-x^(2),x^(2)),(y^(2)-(z+x)^(2),y^(2)-(z+x)^(2),(z+x)^(2)):}|(C_(1)toC_(1)-C_(3),C_(2)toC_(2)-C_(3))`
`=|{:((x+y+z)(x+y-z),0,z^(2)),(0,(z+y+x)(z+y-x),x^(2)),((y+z+x)(y-z-x),(y+z+x)(y-z-x),(z+x)^(2)):}|`
अब ,प्रथम व द्वितीय स्तम्भ में से `(x+y+z)` बाहर लेने पर,
`=(x+y+z)^(2)|{:(x+y-z,0,z^(2)),(0,z+y-x,x^(2)),(-2x,-2z,2xz):}|(R_(3)toR_(3)-(R_(1)+R_(2))`
`=(x+y+z)^(2)|{:((x+y),(z^(2))/(x),z^(2)),((x^(2))/(z),(z+y),x^(2)),(0,0,2xz):}|(C_(1)toC_(1)+(1)/(z)C_(3),C_(2)toC_(2)+(1)/(x)C_(3))`
`=(x+y+z)^(2)[2xz{(x+y)(z+y)-(x^(2))/(z)xx(z^(2))/(x)}]`
`=(x+y+z)^(2)[2xz(xz+xy+yz+y^(2)-xz)]`
`=(x+y+z)^(2)[2xz(xy+yz+y^(2))]`
`=(x+y+z)^(2)[2xzy(x+z+y)]=2xyz(x+y+z)^(3)`