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(C) The given equation is a homogeneous equation of degree $3$,which represents three lines passing through the origin. Let the lines be $y = m_1x$,$y

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(C) The given equation is a homogeneous equation of degree $3$,which represents three lines passing through the origin. Let the lines be $y = m_1x$,$y = m_2x$,and $y = m_3x$.Dividing the equation $2x^3+x^2y+y^3=0$ by $x^3$,we get $2 + \frac{y}{x} + (\frac{y}{x})^3 = 0$.Let $m = \frac{y}{x}$,then $m^3 + m + 2 = 0$.By inspection,$m = -1$ is a root because $(-1)^3 + (-1) + 2 = -1 - 1 + 2 = 0$.Thus,$(m+1)$ is a factor. Dividing $m^3 + m + 2$ by $(m+1)$,we get $m^2 - m + 2 = 0$.The roots of $m^2 - m + 2 = 0$ are $m = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm i\sqrt{7}}{2}$.Since the problem states that two lines are mutually perpendicular,their slopes $m_1$ and $m_2$ must satisfy $m_1m_2 = -1$.However,the roots of the quadratic part are complex. Re-evaluating the equation: if the equation is $x^3 + x^2y - 2y^3 = 0$ (a common variation),the slopes would be real. Given the specific equation $2x^3+x^2y+y^3=0$,the only real slope is $m = -1$.

If two of the lines represented by $2x^3+x^2y+y^3=0$ are mutually perpendicular,then the slope of the third line is

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