TS EAMCET · Maths · Basic of Mathematics
If \(\frac{x^3+3}{(x-3)^3}=a+\frac{b}{x-3}+\frac{c}{(x-3)^2}+\frac{d}{(x-3)^3}\) then \((a+d)-(b+c)=\)
- A \(49\)
- B \(15\)
- C \(-30\)
- D \(-5\)
Answer & Solution
Correct Answer
(D) \(-5\)
Step-by-step Solution
Detailed explanation
Let \(y=x-3 \Rightarrow x=y+3\). \(x^3+3 = (y+3)^3+3 = y^3+9y^2+27y+27+3 = y^3+9y^2+27y+30\). \(\frac{y^3+9y^2+27y+30}{y^3} = 1+\frac{9}{y}+\frac{27}{y^2}+\frac{30}{y^3}\). Comparing with \(a+\frac{b}{y}+\frac{c}{y^2}+\frac{d}{y^3}\), we have \(a=1, b=9, c=27, d=30\).…
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