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Binomial Theorem — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsBinomial Theorem3 Marks
Question
Using binomial theorem write down the expansions of the following:
$\Big(\text{x}+1-\frac{1}{\text{x}}\Big)^3$
✓

Answer

Let $y = x + 1,$ then
$\Big(\text{x}+1-\frac{1}{\text{x}}\Big)^3=\Big(\text{y}-\frac{1}{\text{x}}\Big)^3$
The expansion of $(x + y)^n$ has $n + 1$ terms so the expansion of $\Big(\text{y}-\frac{1}{\text{x}}\Big)^3$ has $4$ terms. Using binomial theorem to expand, we get
$\Big(\text{y}-\frac{1}{\text{x}}\Big)^3={^3\text{C}}_0\text{y}^3\Big( \frac{1}{\text{x}}\Big)^0-{^3\text{C}}_1\text{y}^2\Big(\frac{1}{\text{x}}\Big)+{^3\text{C}}_2\text{y}\Big(\frac{1}{\text{x}}\Big)^2-{^3\text{C}}_3\text{y}^0\Big(\frac{1}{\text{x}}\Big)^3$
$=\text{y}^3-3\text{y}^2\times\frac{1}{\text{x}}+3\text{y}\times\frac{1}{\text{x}^2}-\frac{1}{\text{x}^3}$
Putting $y = x + 1$, we get
$\Big(\text{x}+1-\frac{1}{\text{x}}\Big)^3=(\text{x}+1)^3-3(\text{x}+1)^2\times\frac{1}{\text{x}}+3(\text{x}+1)\times\frac{1}{\text{x}^2}-\frac{1}{\text{x}^3}$
$=\text{x}^3+1+3\text{x}^2+3\text{x}-3\text{x}-\frac{3}{\text{x}}-6+\frac{3}{\text{x}}+\frac{3}{\text{x}^2}-\frac{1}{\text{x}^3}$
$=\text{x}^3+3\text{x}^2-5+\frac{3}{\text{x}^2}-\frac{1}{\text{x}^3}$

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