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1

Ejemplo resuelto de teorema del binomio

$\left(x+3\right)^5$
2

Podemos expandir la expresión $\left(x+3\right)^5$ usando el binomio de Newton, el cual es una fórmula que nos permite obtener la forma expandida de un binomio elevado a un número entero $n$. La fórmula tal cual es: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$
El número de términos que resultan de la expansión es siempre igual a $n+1$. Los coeficientes $\left(\begin{matrix}n\\k\end{matrix}\right)$ son números combinatorios los cuales corresponden a la fila enésima del triángulo de Tartaglia (o triángulo de Pascal). En la fórmula, podemos observar que el exponente de $a$ va disminuyendo, de $n$ a $0$, mientras que el exponente de $b$ va aumentando, de $0$ a $n$. Si uno de los términos del binomio es negativo, se alternan los signos positivos y negativos.

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}3^{0}+\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}3^{1}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
3

Calcular la potencia $3^{0}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}3^{1}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
4

Calcular la potencia $3^{1}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
5

Calcular la potencia $3^{2}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
6

Calcular la potencia $3^{3}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
7

Calcular la potencia $3^{4}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
8

Calcular la potencia $3^{5}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
9

Calcular el coeficiente binomial $\left(\begin{matrix}5\\0\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
10

Calcular el coeficiente binomial $\left(\begin{matrix}5\\1\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
11

Calcular el coeficiente binomial $\left(\begin{matrix}5\\2\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
12

Calcular el coeficiente binomial $\left(\begin{matrix}5\\3\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
13

Calcular el coeficiente binomial $\left(\begin{matrix}5\\4\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
14

Calcular el coeficiente binomial $\left(\begin{matrix}5\\5\end{matrix}\right)$ aplicando la fórmula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
15

Restar los valores $5$ y $-1$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
16

Restar los valores $5$ y $-2$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
17

Restar los valores $5$ y $-3$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
18

Restar los valores $5$ y $-4$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
19

Restar los valores $5$ y $-5$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
20

Sumar los valores $5$ y $0$

$1x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
21

El factorial de $0$ es $1$

$1x^{5}\frac{5!}{1\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
22

El factorial de $5$ es $120$

$1x^{5}\frac{5!}{1\cdot 120}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
23

El factorial de $5$ es $120$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{5!}{\left(1!\right)\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
24

El factorial de $1$ es $1$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{5!}{1\left(4!\right)}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
25

El factorial de $4$ es $24$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{5!}{1\cdot 24}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
26

El factorial de $5$ es $120$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{120}{1\cdot 24}+9x^{3}\frac{5!}{\left(2!\right)\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
27

El factorial de $2$ es $2$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{120}{1\cdot 24}+9x^{3}\frac{5!}{2\left(3!\right)}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
28

El factorial de $3$ es $6$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{120}{1\cdot 24}+9x^{3}\frac{5!}{2\cdot 6}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
29

El factorial de $5$ es $120$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{120}{1\cdot 24}+9x^{3}\frac{120}{2\cdot 6}+27x^{2}\frac{5!}{\left(3!\right)\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
30

El factorial de $3$ es $6$

$1x^{5}\frac{120}{1\cdot 120}+3x^{4}\frac{120}{1\cdot 24}+9x^{3}\frac{120}{2\cdot 6}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
31

Multiplicar $1$ por $120$

$1x^{5}\frac{120}{120}+3x^{4}\frac{120}{1\cdot 24}+9x^{3}\frac{120}{2\cdot 6}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
32

Multiplicar $1$ por $24$

$1x^{5}\frac{120}{120}+3x^{4}\frac{120}{24}+9x^{3}\frac{120}{2\cdot 6}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
33

Multiplicar $2$ por $6$

$1x^{5}\frac{120}{120}+3x^{4}\frac{120}{24}+9x^{3}\frac{120}{12}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
34

Dividir $120$ entre $120$

$1\cdot 1x^{5}+3x^{4}\frac{120}{24}+9x^{3}\frac{120}{12}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
35

Dividir $120$ entre $24$

$1\cdot 1x^{5}+5\cdot 3x^{4}+9x^{3}\frac{120}{12}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
36

Dividir $120$ entre $12$

$1\cdot 1x^{5}+5\cdot 3x^{4}+10\cdot 9x^{3}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
37

Multiplicar $1$ por $1$

$1x^{5}+5\cdot 3x^{4}+10\cdot 9x^{3}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
38

Multiplicar $5$ por $3$

$1x^{5}+15x^{4}+10\cdot 9x^{3}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
39

Multiplicar $10$ por $9$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{5!}{6\left(2!\right)}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
40

El factorial de $2$ es $2$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{5!}{6\cdot 2}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
41

El factorial de $5$ es $120$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{5!}{\left(4!\right)\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
42

El factorial de $4$ es $24$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{5!}{24\left(1!\right)}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
43

El factorial de $1$ es $1$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{5!}{24\cdot 1}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
44

El factorial de $5$ es $120$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{120}{24\cdot 1}+243x^{0}\frac{5!}{\left(5!\right)\left(0!\right)}$
45

El factorial de $5$ es $120$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{120}{24\cdot 1}+243x^{0}\frac{5!}{120\left(0!\right)}$
46

El factorial de $0$ es $1$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{120}{24\cdot 1}+243x^{0}\frac{5!}{120\cdot 1}$
47

El factorial de $5$ es $120$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{6\cdot 2}+81x^{1}\frac{120}{24\cdot 1}+243x^{0}\frac{120}{120\cdot 1}$
48

Multiplicar $6$ por $2$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{12}+81x^{1}\frac{120}{24\cdot 1}+243x^{0}\frac{120}{120\cdot 1}$
49

Multiplicar $24$ por $1$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{12}+81x^{1}\frac{120}{24}+243x^{0}\frac{120}{120\cdot 1}$
50

Multiplicar $120$ por $1$

$1x^{5}+15x^{4}+90x^{3}+27x^{2}\frac{120}{12}+81x^{1}\frac{120}{24}+243x^{0}\frac{120}{120}$
51

Dividir $120$ entre $12$

$1x^{5}+15x^{4}+90x^{3}+10\cdot 27x^{2}+81x^{1}\frac{120}{24}+243x^{0}\frac{120}{120}$
52

Dividir $120$ entre $24$

$1x^{5}+15x^{4}+90x^{3}+10\cdot 27x^{2}+5\cdot 81x^{1}+243x^{0}\frac{120}{120}$
53

Dividir $120$ entre $120$

$1x^{5}+15x^{4}+90x^{3}+10\cdot 27x^{2}+5\cdot 81x^{1}+1\cdot 243x^{0}$
54

Multiplicar $10$ por $27$

$1x^{5}+15x^{4}+90x^{3}+270x^{2}+5\cdot 81x^{1}+1\cdot 243x^{0}$
55

Multiplicar $5$ por $81$

$1x^{5}+15x^{4}+90x^{3}+270x^{2}+405x^{1}+1\cdot 243x^{0}$
56

Multiplicar $1$ por $243$

$1x^{5}+15x^{4}+90x^{3}+270x^{2}+405x^{1}+243x^{0}$
57

Cualquier expresión elevada a la potencia uno es igual a esa misma expresión

$1x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243x^{0}$
58

Cualquier expresión algebraica multiplicada por uno es igual a esa misma expresión

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243x^{0}$
59

Cualquier expresión matemática elevada a la potencia $0$ es igual a $1$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243\cdot 1$
60

Multiplicar $243$ por $1$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

Respuesta Final

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

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