Prove that `|(x,p,q),(p,x,q),(p,q,x)| = (x - p)(x - q)(x + p +q)`

Show that |(x,p,q),(p,x,x),(q,q,x)| = (x - p)(x^(2) + px - 2q^2)

The value of {:[(x,p,q),(p,x,q),(p,q,x)]:} is

If x^(pq) = (x^(p))^(q) , then p =

(p ^^ ~ q) ^^(~ p vv q) is

The roots of the equation (q- r) x^(2) + (r - p) x + (p - q)= 0 are

(p ^^ ~q)^^(~p ^^ q) is

Add: p(p-q),q(q-r) and r(r-p)

(p^^~q)vv(~p^^q) is :

Using Properties of determinants, prove that: {:|(b+c,c+a,a+b),(q+r,r+p,p+q),(y+z,z+x,x+y)|=2{:|(a,b,c),(p,q,r),(x,y,z)|

int dx/((x-p)sqrt((x-p)(x-q)) =