By using binomial theorem, expand $(1 + x + x^{2})^{3}$ .

Open in App

Solution

Let $y = x + x^{2}$
Then, $(1 + x + x^{2})^{3}$
$= (1 + y)^{3} =^{3} C_{0} +^{3} C_{1} +^{3} C_{1} +^{3} C_{2} y^{2} +^{3} C_{3} y^{3}$
$= 1 + 3 y + 3 y^{2} + y^{3}$
$= 1 + 3 (x + x^{2}) + 3 (x + x^{2})^{2} + (x + x^{2})^{3}$
$= 1 + 3 (x + x^{2}) + 3 (x^{2} + 2 x^{3} + x^{4}) + {^{3} C_{0} x^{3} x^{2})^{0} +^{3} C_{1} (x^{2}) x^{2})^{1} +^{3} C_{2} x^{1} (x^{2})^{2} +^{3} C_{3} (x^{2})^{3}}$
$= 1 + 3 (x + x^{2}) + 3 (x^{2} + 2 x^{3} + x^{4}) + (x^{3} + 3 x^{4} + 3 x^{5} + x^{6})$
$= x^{6} + 3 x^{5} + 6 x^{4} + 7 x^{3} + 6 x^{2} + 3 x + 1$ .

Suggest Corrections

Similar questions

Expand $(1 + x + x^{2})^{3}$ using binomial expansion.

{(1 + \frac{x}{2} - \frac{2}{x})}^{4}, x \neq 0

{(1 + \frac{x}{2} - \frac{2}{x})}^{4}, x \neq 0

{(1 + \frac{x}{2} - \frac{2}{x})}^{4}, x \neq 0

{[1 + \frac{x}{2} - \frac{2}{x}]}^{4}, x \neq 0

Join BYJU'S Learning Program