If the equation y^3 - 3x^2y + m(x^3 - 3xy^2) | vedclass

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(A) The given equation is $y^3 - 3xy^2m + 3x^2y(-1) - mx^3 = 0$. Dividing by $x^3$,we get $(y/x)^3 - 3m(y/x)^2 - 3(y/x) - m = 0$. Let $t = y/x = 	an

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Lines are equally inclined to each other

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(A) The given equation is $y^3 - 3xy^2m + 3x^2y(-1) - mx^3 = 0$. Dividing by $x^3$,we get $(y/x)^3 - 3m(y/x)^2 - 3(y/x) - m = 0$. Let $t = y/x = 	an 	heta$. Then $t^3 - 3mt^2 - 3t - m = 0$. This is a cubic equation in $t$. If we substitute $t = 	an 	heta$,the equation represents three lines $y = t_1x, y = t_2x, y = t_3x$. It can be shown that the angles between these lines are $\pi/3$ or $60^\circ$. Thus,the lines are equally inclined to each other.

If the equation $y^3 - 3x^2y + m(x^3 - 3xy^2) = 0$ represents three lines passing through the origin,then:

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