Latest CTS Aptitude Question SOLUTION: .if p+q+r=0 then p^2/qr+q^2/pr+r^2/pq=?

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.if p+q+r=0 then p^2/qr+q^2/pr+r^2/pq=?

Asked In CTS amit shekhar (11 years ago)

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Pallavi Barjatiya  (11 years ago)

ans is 3!
we know the relation if p+Q+R =0 then P3+q3+r3 = 3pqr
here we can also write the given problem in the form: (p^3+q^3+r^3)/pqr = 3pqr/pqr= 3. ans

Is this solution Helpfull? Yes (16) | No (11)  

sai spandana  (11 years ago)

answer is zero let us have the lcm we have p^3+q^3+r^3/p*q*r since we have a^3+b^3+c^3=(a+b+c)(......) since given p+q+r=0 ans is zero

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gopal kumar  (11 years ago)

3 is answer

Is this solution Helpfull? Yes (2) | No   

sakthivel  (11 years ago)

wheather any relationship between p,q & r
in case we consider p=1 q=2 r=-3
p^2/qr+q^2/pr+r^2/pq= 15/6
ie the case and ans will depend upon the condition or option

Is this solution Helpfull? Yes (1) | No (5)  

venkatanarayana pamidi  (9 years ago)

p3+q3+r3/pqr==>(p+q+r)(2pq+2qr+2rp)/pqr==>0

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Ananda Mohan Chakraborty  (7 years ago)

as simplyfying p2/qr+q2/pr+r2/pq we get
(p3+q3+r3)/pqr
as a3+b3+c3=(a+b+c) (a2+b2+c2-ab-bc-ca)+3abc
we know a+b+c=0
hence a3+b3+c3=3abc
ie; 3pqr/pqr=3
ans is 3

Is this solution Helpfull? Yes (1) | No   

suryacsr  (6 years ago)

let us assume q+r be s then do cube
(p+s)^3=p^3+s^3+3p^2s+3ps^2
substitute assumption s equation

(p+q+r)^3= p^3 + (q+r)^3 + 3p^2(q+r) +3p(q+r)^2

In the question they give p+q+r=0

0=p^3 + q^3+r^3 +(3q^2r +3r^2q+3p^2r+3p^2q+3pq^2+3pr^2+6pqr)

inside braces equation get common 3pqr
0=p^3+q^3+r^3+3pqr{2(p+q+r)+2 }

again apply p+q+r=0

0=p^3+q^3+r^3+3pqr(2)
-6pqr=p^3+q^3+r^3

-6=p^2/qr+q^2/pr+r^2/pq


therefore -6 is the answer

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