Question

The locus of points $(x,y)$, for which $x^3+y^3+3xy-1=0$ will be

• A. A line
• B. A line and a point
• C. A line and 2 point
• D. None of these

Answer 1

The correct option is B A line and a point

The identity in the passage is
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
So, we write the given equation (in the question) as $x^3+y^3+(-1{)}^3=-3xy$
ie, $x^3+y^3+(-1{)}^3=3(x\left)\right(y\left)\right(-1)$
Thus if we put $a=x,b=y,c=-1$, we get
$a^3+b^3+c^3=3abc$
Hence, $a+b+c=0$; $a^2+b^2+c^2-ab-bc-ca=0$
ie., $(a+b+c=0)$; $\frac{1}{2}\left(\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2)=0$
$\Rightarrow x+y-1=0$; $\frac{1}{2}\left[\right(x-y{)}^2+(y+1{)}^2+(-1-x{)}^2]=0$
ie, $x+y=1$ A line and
$x=y=-1$ A point (-1,-1)