Show that : `(4p q+3q)^2-(4q p-3q)^2=48 p q^2` `(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0`

Show that. (i) (3x+7)^2-84x=(3x-7)^2 (ii) (9p-5q)^2+180pq=(9p+5q)^2 (iii) (4/3(m)-3/4(n))^2+2mn=16/9(m^2)+9/16(n^2) (iv) (4pq+3q)^2-(4pq-3q)^2=48pq^2 (v) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

If pth, qth, and rth terms of an A.P. are a ,b ,c , respectively, then show that (1) a(q-r)+b(r-p)+c(p-q)=0 (2)(a-b)r+(b-c)p+(c-a)q=0

Let A=[{:(a,b,c),(p,q,r),(1,1,1):}]and B=A^(2) If (a-b)^(2) +(p-q)^(2) =25, (b-c) ^(2)+ (q-r)^(2)= 36 and (c-a)^(2) +(r-p)^(2)=49, then det B is

If A be one A.M. and p ,q be two G.M. ' s between two numbers, then 2A is equal to (a) (p^3+q^3)/(p q) (b) (p^3-q^3)/(p q) (c) (p^2+q^2)/2 (d) (p q)/2

(p ^^ ~ q)vv q is logical equivalent to (a) p vv q (b) ~P (c) p (d) ~q

The common difference of the A.P. is 1/(2q),(1-2q)/(2q),(1-4q)/(2q),\ dot is (a)-1 (b) 1 (c) q (d) 2q

The base B C of a A B C is bisected at the point (p ,q) & the equation to the side A B&A C are p x+q y=1 & q x+p y=1 . The equation of the median through A is: (p-2q)x+(q-2p)y+1=0 (p+q)(x+y)-2=0 (2p q-1)(p x+q y-1)=(p^2+q^2-1)(q x+p y-1) none of these

The sums of first p, q, r terms of an A.P. are a, b, c respectively. Prove that (a)/(p) (q-r) +(b)/(q) (r-p) +(c )/(r) (p-q) =0

If the equation x^2-3p x+2q=0a n dx^2-3a x+2b=0 have a common roots and the other roots of the second equation is the reciprocal of the other roots of the first, then (2-2b)^2 . a. 36p a(q-b)^2 b. 18p a(q-b)^2 c. 36b q(p-a)^2 d. 18b q(p-a)^2

Show 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|