Factorize : `p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3`

Prove that p x^(q-r)+q x^(r-p)+r x^(p-q)> p+q+r ,where p, q, r are distinct and x!=1.

If zeros of the polynomial f(x)=x^3-3p x^2+q x-r are in A.P., then (a) 2p^3=p q-r (b) 2p^3=p q+r (c) p^3=p q-r (d) None of these

If q is the mean proportional between p and r , show that : pqr (p + q + r)^(3) = (pq + qr + pr)^(3) .

If omega is an imaginary cube root of unity, then the value of (p+q)^3+(pomega+qomega^2)^3+(pomega^2+qomega)^3 is: (a) 3(p+q)(p+qomega)(p+qomega^2) (b) 3(p^3+q^3) (c) 3(p^3+q^3)-p q(p+q) (d) 3(p^3+q^3)p q(p+q)

If (x+2) is a common factor of (px^2+qx+r) and (qx^2+px+r) then (a) p=q or p+q+r=0 (b) p=r or p+q+r=0 (c) q=r or p+q+r=0 (d) p=q=-1/2r

If t_n denotes the nth term of an A.P. and t_p= 1/q, t_q= 1/p then which one of the following is necessarily a root of the equation (p+2q-3r)x^2+(q+2r-3p)x+(r+2p-3q)=0 (A) t_p (B) t_q (C) t_(pq) (D) t_(p+q)

The roots of the equation (q-r)x^2+(r-p)x+p-q=0 are (A) (r-p)/(q-r),1 (B) (p-q)/(q-r),1 (C) (q-r)/(p-q),1 (D) (r-p)/(p-q),1

Without expanding the determinant prove that: |(0, p-q, p-r), (q-p,0, q-r),(r-p, r-q, 0)|=0

Let p and q be real numbers such that p!=0,p^3!=q ,and p^3!=-qdot If alpha and beta are nonzero complex numbers satisfying alpha+beta=-p and alpha^3+beta^3=q , then a quadratic equation having alpha//beta and beta//alpha as its roots is A. (p^3+q)x^2-(p^3+2q)x+(p^3+q)=0 B. (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 C. (p^3+q)x^2-(5p^3-2q)x+(p^3-q)=0 D. (p^3+q)x^2-(5p^3+2q)x+(p^3+q)=0

If (logx)/(q-r)= (logy)/(r-p)=(logz)/(p-q) prove that x^(q+r).y^(r+p).z^(p+q)=x^p.y^q.z^r