If frac{x^3+3}{(x-3)^3}=a+frac{b}{x-3}+frac{c | vedclass

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(D) Let $x-3 = y$, so $x = y+3$. Substituting this into the expression: $\frac{(y+3)^3+3}{y^3} = \frac{y^3+9y^2+27y+27+3}{y^3} = \frac{y^3+9y^2+27y+30

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$-5$

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(D) Let $x-3 = y$, so $x = y+3$. Substituting this into the expression: $\frac{(y+3)^3+3}{y^3} = \frac{y^3+9y^2+27y+27+3}{y^3} = \frac{y^3+9y^2+27y+30}{y^3} = 1 + \frac{9}{y} + \frac{27}{y^2} + \frac{30}{y^3}$. Comparing this with $a + \frac{b}{y} + \frac{c}{y^2} + \frac{d}{y^3}$, we get: $a = 1, b = 9, c = 27, d = 30$. Now, calculate $(a+d)-(b+c)$: $(1+30) - (9+27) = 31 - 36 = -5$.

If $\frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3}$ then $(a+d)-(b+c)=$

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