1x3+y3+3xy-1 =0
=> (x+y)3 - 3xy(x+y) +3xy -1 =0
=> (x+y)3-1 = 3xy(x+y) - 3xy
=> (x+y-1)(x2+y2+2xy +1 +x +y) = 3xy(x+y-1)
=> (x+y-1)[x2+y2 - xy +1 +x +y] = 0
12 Answers
1is it C....
but i am getting a line and a crve[which is not a circle]......
21sky, pl. give da the equation of ur line n ur curve
it is not da rite answer though
1
33that 2 degree... is a point (-1,-1)
solve it as a quadratic and D≤0
so D=0 for x-=-1 y=-1
21HAAA abhi yaad aaya!!!
ther is a way to find da exact pt. par uski zaroorat nahi hai
C this u'll get how :
33This is one of the example of some must see problems...Q.3
http://targetiit.com/iit_jee_forum/posts/some_must_see_problems_155.html
33hahah :D
utna mat yaad rakho...
they are just second degree.. solve it as quadratic.... and interpret from there....
21ANOTHER methd :
[x^2+y^2 - xy +1 +x +y] has to be 0
and v can c that the eqtn is symmetric wid x and y
so it bcums
x^ + 2x + 1 = (x+1)2 = 0
ther4 x = -1 = y
thnx PRiyam n Sky [1]
33hmm.... yes.. is question me symmetry tah dekhe hi nahi... [3]