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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 21 of 160

Marks: +1, -0

If

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

(a+d)-(b+c)=

49
15
-30
-5

Solution:

Given,

\;\frac{x^3+3}{(x-3)^3}=a+\;\frac{b}{x-3}+\;\frac{c}{(x-3)^2}+\;\frac{d}{(x-3)^3} \dots(i)

Let

y=x-3

\Rightarrow x=y+3

Then,

\;\frac{x^3+3}{(x-3)^3}=\;\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}

\;=1+\;\frac{9}{x-3}+\;\frac{27}{(x-3)^2}+\;\frac{30}{(x-3)^3}

Comparing this with Eq. (i), we get

a=1, b=9, c=27, d=30

So,

(a+d)-(b+c)=(1+30)-(9+27)

=31-36=-5

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