Question

If $x+y$ and $2x-y$ are factors of $\lambda x^3-x^{2}y+\mu x{y}^2+y^3,x,y\ne 0$,

• A. $(-2,2)$
• B. $(2,2)$
• C. $(2,-2)$
• D. $noneofthese$

Answer 1

The correct option is A $(-2,2)$

$\because (x+y)$ and $(2x-y)$ are factors of $\lambda x^3-x^2\gamma +\mu x{y}^2+y^3----\left(1\right)$

$\therefore$ The $y=0\Rightarrow x=-y$ and $2x-y=0\Rightarrow y=2x$

So, for $y=-x$ and $y=2x,$ from $\left(1\right)$,

$\lambda x^3-x^2(-x)+\mu x(-x{)}^2+(-x{)}^3=0$

or $\lambda x^3+a^3\mu x^3-x^3=0$

or, $\lambda +\mu =0[\because x\ne 0]$

$\therefore \lambda =-\mu$

$\lambda x^3-x^2\left(2x\right)+\mu x(2x{)}^2+(2x{)}^3=0$

or $\lambda x^3-2x^3+4\mu x^3+2x^3=0$

or $\lambda -6+4\mu =0\Rightarrow -\mu -6+4\mu =0$

$\therefore \lambda =-2$ or $-6+3\mu =0\Rightarrow \mu =2$

$\therefore$ Required solution $(-2,2)$