If the determinant `|(x+a,p+u,l+f),(y+b,q+v,m+g),(z+c,r+w,n+h)|` splits into exactly `k` determinants of order 3, each element of which contains onlyone term, then the value of `k` is 8.

Answer

N/a

Explanation

True
Since `|(x+a,p+u,l+f),(y+b,q+v,m+g),(z+c,r+w,n+h)|`
`=|(x,p,l),(y+b,q+v,m+g),(z+c,r+w,n+h)|+|(a,u,f),(y+b,q+v,m+g),(z+c,r+w,n+h)|` [splitting first row]
`=|(x,p,l),(y,q,m),(z+c,r+m,n+h)|+|(x,p,l),(b,v,g),(z+c,r+w,n+h)|`
`+|(a,u,f),(y,q,m),(z+c,r+w,n+h)|+|(a,u,f),(b,v,g),(z+c,r+w,n+h)|` [splitting second row]
Similarly we can split these 4 determinants in 8 determinantes by splitting each one in two determinants further. So, given statement is true.