Binomial Theorem — MATHS STD 11 Science — Question

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

$\Big(\text{x}+1-\frac{1}{\text{x}}\Big)^3=(\text{x}+1)^3-3(\text{x}+1)^2\times\frac{1}{\text{x}}+3(\text{x}+1)\times\frac{1}{\text{x}^2}-\frac{1}{\text{x}^3}$

$=\text{x}^3+1+3\text{x}^2+3\text{x}-3\text{x}-\frac{3}{\text{x}}-6+\frac{3}{\text{x}}+\frac{3}{\text{x}^2}-\frac{1}{\text{x}^3}$

$=\text{x}^3+3\text{x}^2-5+\frac{3}{\text{x}^2}-\frac{1}{\text{x}^3}$