Solve:-
`(p^2q^2r^5 - p^3q^5r^3 + p^5 q^3r^3) div p^2q^2r^2`

Solve:- (p^2q^2r^4 - p^3q^5r^3 + p^5q^3r^2) div p^2q^2r^2

Let's resolve into factors: (p^2 - 3q^2)^2 - 16 (p^2 - 3q^2) + 63

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 p=a+bomega+comega^2 , q=b+comega+aomega^2 , and r=c+aomega+bomega^2 , where a ,b ,c!=0 and omega is the complex cube root of unity, then (a) p+q+r=a+b+c (b) p^2+q^2+r^2=a^2+b^2+c^2 (c) p^2+q^2+r^2=-2(p q+q r+r p) (d) none of these

Prove that px^(q - r) + qx^(r - p) + rx^(p - q) gt p + q + r where p,q,r are distinct number and x gt 0, x =! 1 .

Let p , q , r in R^(+) and 27pqr ge (p+q+r)^(3) and 3p+4q+5r=12 . Then the value of 8p+4q-7r=

If p=sin(A-B)sin(C-D),q=sin(B-C)sin(A-D) , r=sin(C-A)sin(B-D) then (a) p+q-r=0 (b) p+q+r=0 p-q+r=0 (d) p^3+q^3+r^3=3p q r

Let's express the following in perfect square and hence find the value. p^2q^2 + 10pqr + 25r^2 When p = 2 ,q = -1, r = 3

Prove that the equation p cos x - q sin x = r admits solution for x if and only if -sqrt(p^2 + q^2) lt r lt sqrt(p^2 + q^2)

If one root is square of the other root of the equation x^2+p x+q=0, then the relation between pa n dq is p^3-q(3p-1)+q^2=0 p^3-q(3p+1)+q^2=0 p^3+q(3p-1)+q^2=0 p^3+q(3p+1)+q^2=0