Question

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

Answer 1

Hint – In this question let $\left( {x + {x^2}} \right)$ be equal to some variable, so that the given expression
Given expression can be converted to the standard binomial expansion form ${\left( {1 + x} \right)^n}$, whose expansion according to Binomial theorem is given as, ${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + .............. + {}^n{C_n}{x^n}$.

Complete step-by-step answer:
Given expression
${\left( {1 + x + {x^2}} \right)^3}$
Let
$t = x + {x^2}$
$\Rightarrow {\left( {1 + x + {x^2}} \right)^3} = {\left( {1 + t} \right)^3}$
Now as we know according to Binomial theorem the expansion of ${\left( {1 + x} \right)^n}$ is
$\Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + .............. + {}^n{C_n}{x^n}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property in above equation we have,
$\Rightarrow {}^n{C_0} = \dfrac{{n!}}{{0!\left( {n - 0} \right)!}} = 1$, ${}^n{C_1} = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}} = n$, ${}^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = \dfrac{{n\left( {n - 1} \right)}}{{2!}}$ , ${}^n{C_3} = \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}$ and so on......., so the above equation converts into,
$\Rightarrow {\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ..............$
Therefore the expansion of ${\left( {1 + t} \right)^3}$according to binomial theorem is
$\Rightarrow {\left( {1 + t} \right)^3} = 1 + 3t + \dfrac{{3\left( {3 - 1} \right)}}{{2!}}{t^2} + \dfrac{{3\left( {3 - 1} \right)\left( {3 - 2} \right)}}{{3!}}{t^3}$
Now simplify the above equation we have,
$\Rightarrow {\left( {1 + t} \right)^3} = 1 + 3t + \dfrac{{3\left( 2 \right)}}{{2 \times 1}}{t^2} + \dfrac{{3\left( 2 \right)\left( 1 \right)}}{{3 \times 2 \times 1}}{t^3}$
$\Rightarrow {\left( {1 + t} \right)^3} = 1 + 3t + 3{t^2} + {t^3}$
Now re substitute the value of (t) we have,
$\Rightarrow {\left( {1 + x + {x^2}} \right)^3} = 1 + 3\left( {x + {x^2}} \right) + 3{\left( {x + {x^2}} \right)^2} + {\left( {x + {x^2}} \right)^3}$
Now expand the square and cube according to property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b$ so we have,
$\Rightarrow {\left( {1 + x + {x^2}} \right)^3} = 1 + 3\left( {x + {x^2}} \right) + 3\left( {{x^2} + {x^4} + 2{x^3}} \right) + \left( {{x^3} + {x^6} + 3{x^5} + 3{x^4}} \right)$
Now simplify the above equation we have,
$\Rightarrow {\left( {1 + x + {x^2}} \right)^3} = 1 + 3x + 3{x^2} + 3{x^2} + 3{x^4} + 6{x^3} + \left( {{x^3} + {x^6} + 3{x^5} + 3{x^4}} \right)$
$\Rightarrow {\left( {1 + x + {x^2}} \right)^3} = 1 + 3x + 6{x^2} + 7{x^3} + 6{x^4} + 3{x^5} + {x^6}$
So this is the required expansion of a given expression using the Binomial theorem.
So this is the required answer.

Note – Binomial theorem specifies the expansion of any power ${\left( {a + b} \right)^m}$ of a binomial $\left( {a + b} \right)$ as a sum of products ${a^i}{b^j}$, such as ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Important point here is that every algebraic standard identity is derived from the same concept as being explained.