Correct Answer - D
The equation `x^(3)+y^(3)+3xy=1` can be written as
`(x+y)^(2)-3xy(x+y)+3xy-1=0`
`rArr {(x+y)^(3)-1^(3)}-3xy(x+y-1)=0`
`rArr (x+y-1)(x^(2)+2xy+y^(2)+x+y+1)-3xy(x+y-1)=0`
`rArr (x+y-1)(x^(2)+y^(2)-xy+x+y+1)=0`
`rArr x+y-1=0, x^(2)+y^(2)-xy+x+y+1=0`
Comparing equation `x^(2)+y^(2)-xy+x+y+1=0` with the equation `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`, we get
a=1, b=1, c=1, `h=-(1)/(2), g=(1)/(2), f=(1)/(2)`
`:. abc+2fgh-af^(2)-bg^(2)-ch^(2)=1-(1)/(4)-(1)/(4)-(1)/(4)-(1)/(4)=0`
So, `x^(2)+y^(2)-xy+x+y+1=0` represents a pair of straight lines.
Hence, statement-1 is not true. However, statement-2 is true.