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# Find the coefficient of ${{\text{x}}^3}{\text{ in }}{\left( {1 + {\text{x + }}{{\text{x}}^2}} \right)^3}.$

Last updated date: 19th Mar 2023
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Hint: For finding the coefficient of ${{\text{x}}^3}$, we have to first expand the given algebraic expression. The given algebraic expression is in the form of ${\left( {{\text{a + b + c}}} \right)^3}$. So to get the term of different powers, we have to expand this expression. After expanding, collect the terms with power 3.

In the question, we have to find the coefficient of ${{\text{x}}^3}$. The algebraic expression given is:
${\left( {1 + {\text{x + }}{{\text{x}}^2}} \right)^3}.$
Now to get the coefficient of ${{\text{x}}^3}$, we have to first expand the given expression.
The given algebraic expression is in the form of ${\left( {{\text{a + b + c}}} \right)^3}$ and we know that:
${\left( {{\text{a + b + c}}} \right)^3} = {{\text{a}}^3} + {{\text{b}}^3} + {{\text{c}}^3} + 3\left( {{\text{a + b}}} \right)\left( {{\text{b + c}}} \right)\left( {{\text{c + a}}} \right).$
${(1 + {\text{x + }}{{\text{x}}^2})^3} = {1^3} + {{\text{x}}^3} + {({{\text{x}}^2})^3} + 3\left( {{\text{1 + x}}} \right)\left( {{\text{x + }}{{\text{x}}^2}} \right)\left( {{{\text{x}}^2}{\text{ + 1}}} \right)$
${(1 + {\text{x + }}{{\text{x}}^2})^3} = {1^3} + {{\text{x}}^3} + {{\text{x}}^6} + 3\left( {{\text{1 + x}}} \right)\left( {{{\text{x}}^4} + {{\text{x}}^3} + {{\text{x}}^2} + {\text{x}}} \right) \\ {(1 + {\text{x + }}{{\text{x}}^2})^3} = 1 + {{\text{x}}^3} + {{\text{x}}^6} + 3\left( {{{\text{x}}^5} + 2{{\text{x}}^4} + 2{{\text{x}}^3} + 2{{\text{x}}^2} + {\text{x}}} \right) \\ {(1 + {\text{x + }}{{\text{x}}^2})^3} = 1 + {{\text{x}}^3} + {{\text{x}}^6} + 3{{\text{x}}^5} + 6{{\text{x}}^4} + 6{{\text{x}}^3} + 6{{\text{x}}^2} + 3{\text{x}} \\ {(1 + {\text{x + }}{{\text{x}}^2})^3} = {{\text{x}}^6} + 3{{\text{x}}^5} + 6{{\text{x}}^4} + 7{{\text{x}}^3} + 6{{\text{x}}^2} + 3{\text{x + 1}}{\text{.}} \\ {\text{So, the final expression that we get is}}: \{{(1 + {\text{x + }}{{\text{x}}^2})^3} = {{\text{x}}^6} + 3{{\text{x}}^5} + 6{{\text{x}}^4} + 7{{\text{x}}^3} + 6{{\text{x}}^2} + 3{\text{x + 1}}$----- (1)
The coefficient of ${{\text{x}}^3}$ is 7.