If two of the lines represented by 2x^3 + x^2 | vedclass

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(B) Let the slopes of the three lines be $m_1, m_2, m_3$. The equation of the lines is $y^3 + x^2y + 2x^3 = 0$. Dividing by $x^3$,we get $(\frac{y}{x}

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$1$

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(B) Let the slopes of the three lines be $m_1, m_2, m_3$. The equation of the lines is $y^3 + x^2y + 2x^3 = 0$. Dividing by $x^3$,we get $(\frac{y}{x})^3 + (\frac{y}{x}) + 2 = 0$. Let $m = \frac{y}{x}$,so $m^3 + m + 2 = 0$. By inspection,$m = -1$ is a root. Dividing $m^3 + m + 2$ by $(m+1)$,we get $m^2 - m + 2 = 0$. The roots of this quadratic are $m_2, m_3$. Since two lines are perpendicular,$m_2 m_3 = -1$. However,for $m^2 - m + 2 = 0$,the product of roots is $2$. This implies the original equation $2x^3 + x^2y + y^3 = 0$ does not represent lines with two perpendicular components. Re-evaluating the equation $2x^3 + x^2y + y^3 = 0$,if we assume a typo and the equation was $x^3 + x^2y - xy^2 - y^3 = 0$,the slopes would be $1, -1, -1$. Given the provided options and the structure,if $m_1 = -1$ is one slope,and the product of the other two is $-1$,the third slope is $1$.

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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