Question

If a,b and c are non zero integers and a+b+c=0. PT a²/bc + b²/ca + c²/ab = 3

Answer 1

Given,

a+b+c=0 and a, b,c are non zero integers.

R.T.P: a²/bc + b²/ca + c²/ab = 3

consider the L.H.S part of the equation,

a²/bc + b²/ca + c²/ab

=a³+b³+c³/abc........(1), we will come back to this part later...

We know that,

a³+b³+c³=(a+b+c)(a²+b²+c²-ab-bc-ca)+3abc

since a+b+c=0

a³+b³+c³=3abc

from (1)

3abc/abc=3=R.H.S

i.e a²/bc + b²/ca + c²/ab = 3

Hence proved

Answer 2

Step-by-step explanation:

it is given,

a+b+c=0

(a+b) = -c

(a+b) ^3=(-c) ^3

now

(a^2/bc) +(b^2/ca) +(c^2/ab)

taking l. c. m

(a^3+b^3+c^3) /abc

(a^3+b^3-(a+b) ^3) /abc

(a^3+b^3-{a^3+b^3+3ab^2+3a^2b})/-ab(a+b)

3ab(b+a)/(ab(a+b))=3