If x = frac{1 + t}{t^3} and y = frac{3}{2t^2} | vedclass

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(C) Given $x = \frac{1}{t^3} + \frac{1}{t^2}$ and $y = \frac{3}{2t^2} + \frac{2}{t}$.Differentiating $x$ with respect to $t$:$\frac{dx}{dt} = -\frac{3

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(C) Given $x = \frac{1}{t^3} + \frac{1}{t^2}$ and $y = \frac{3}{2t^2} + \frac{2}{t}$.Differentiating $x$ with respect to $t$:$\frac{dx}{dt} = -\frac{3}{t^4} - \frac{2}{t^3} = -\frac{3 + 2t}{t^4}$.Differentiating $y$ with respect to $t$:$\frac{dy}{dt} = \frac{3}{2} \left( -\frac{2}{t^3} \right) - \frac{2}{t^2} = -\frac{3}{t^3} - \frac{2}{t^2} = -\frac{3 + 2t}{t^3}$.Now,find $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-(3 + 2t)/t^3}{-(3 + 2t)/t^4} = \frac{t^4}{t^3} = t$.Substitute $\frac{dy}{dx} = t$ into the expression $x \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx}$:$x(t)^3 - t = \left( \frac{1 + t}{t^3} \right) t^3 - t = (1 + t) - t = 1$.

If $x = \frac{1 + t}{t^3}$ and $y = \frac{3}{2t^2} + \frac{2}{t}$,then $x \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx}$ is equal to (where $t$ is a real parameter).

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