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(A) Given the equation $x^4+3x^3-6x^2+2x-4=0$ with roots $\alpha, \beta, \gamma, \delta$. To find the equation with roots $\frac{1}{\alpha}, \frac{1}{

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$4x^4-2x^3+6x^2-3x-1=0$

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(A) Given the equation $x^4+3x^3-6x^2+2x-4=0$ with roots $\alpha, \beta, \gamma, \delta$. To find the equation with roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$,we replace $x$ with $\frac{1}{x}$ in the original equation: $(\frac{1}{x})^4 + 3(\frac{1}{x})^3 - 6(\frac{1}{x})^2 + 2(\frac{1}{x}) - 4 = 0$ Multiply the entire equation by $x^4$: $1 + 3x - 6x^2 + 2x^3 - 4x^4 = 0$ Rearranging the terms: $-4x^4 + 2x^3 - 6x^2 + 3x + 1 = 0$ Multiplying by $-1$: $4x^4 - 2x^3 + 6x^2 - 3x - 1 = 0$

If $\alpha, \beta, \gamma$ and $\delta$ are the roots of the equation $x^4+3x^3-6x^2+2x-4=0$,then find the equation having roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ and $\frac{1}{\delta}$.

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