Question

If p,q and r are all non zero and p+q+r=0 prove that p^2÷qr+q^2÷rp+r^2÷pq=3​

Answer 1

Answer: This question can be easily solved.

Step-by-step explanation:

p+q+r=0

Given that,  

p+q=-r .........(1)

Cubing both sides,

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

p^3+q^3+3pq^2+3p^2q= r^3

p^3+q^3+r^3= -3pq(p+q)

p^3+q^3+r^3= 3pqr

p^3/pqr+q^3/pqr+r^3/pqr= 3

p^2/pqr+q^2/pqr+r^2/pqr= 3

Hence it is proved.

Answer 2

$\dfrac{p^2}{qr}+\dfrac{q^2}{rp}+\dfrac{r^2}{pq}=3$, proved.

Step-by-step explanation:

Given,

p + q + r = 0

To prove that, $\dfrac{p^2}{qr}+\dfrac{q^2}{rp}+\dfrac{r^2}{pq}=3$.

L.H.S. $= \dfrac{p^2}{qr}+\dfrac{q^2}{rp}+\dfrac{r^2}{pq}$

Multiplying p, q and r first, second and third term respectively,

$= \dfrac{p^3}{pqr}+\dfrac{q^3}{pqr}+\dfrac{r^3}{pqr}$

$= \dfrac{1}{pqr}(p^3+q^3+r^3)$

Using the algebraic identity,

If a + b + c = 0, then

$a^{3} +b^{3} +c^{3} =3abc$

$= \dfrac{1}{pqr}(3pqr)$

[∵ p + q + r = 0 given by question]

= 3

= R.H.S., proved.

Thus, $\dfrac{p^2}{qr}+\dfrac{q^2}{rp}+\dfrac{r^2}{pq}=3$, proved.