developpe (x+3)^3 — solution étape par étape | Explain MAths

developpe (x+3)^3

Développons :

{(x + 3)}^{3}

En utilisant la formule du binôme de Newton :

{(a + b)}^{n} = \sum_{k = 0}^{n} (\binom{n!}{k! (n - k)!}) a^{n - k} b^{k}

, on développe l'expression mathématique :

\frac{3!}{3! \cdot 0!} \cdot x^{3} \cdot 3^{0}

+ \frac{3!}{2! \cdot 1!} \cdot x^{2} \cdot 3^{1}

+ \frac{3!}{1! \cdot 2!} \cdot x^{1} \cdot 3^{2}

+ \frac{3!}{0! \cdot 3!} \cdot x^{0} \cdot 3^{3}

ℹ

3^{0}

=

1

‖

\frac{3!}{3! \cdot 0!} \cdot x^{3} \cdot 1

=

\frac{3!}{3! \cdot 0!} \cdot x^{3}

‖

3!

=

6

‖

2!

=

2

‖

1!

=

1

‖

2 \cdot 1

=

2

‖

3^{1}

=

3

‖

3!

=

6

‖

1!

=

1

‖

2!

=

2

‖

1 \cdot 2

=

2

‖

x^{1}

=

x

‖

3^{2}

=

9

‖

x^{0}

=

1

‖

\frac{3!}{0! \cdot 3!} \cdot 1 \cdot 3^{3}

=

\frac{3!}{0! \cdot 3!} \cdot 3^{3}

\frac{3!}{3! \cdot 0!} \cdot x^{3}

+ \frac{6}{2} \cdot x^{2} \cdot 3

+ \frac{6}{2} \cdot x \cdot 9

+ \frac{3!}{0! \cdot 3!} \cdot 3^{3}

ℹ

3!

=

6

‖

3!

=

6

‖

0!

=

1

‖

6 \cdot 1

=

6

‖

\frac{6}{2} \cdot 3

=

\frac{18}{2}

‖

\frac{6}{2} \cdot 9

=

\frac{54}{2}

‖

3!

=

6

‖

0!

=

1

‖

3!

=

6

‖

1 \cdot 6

=

6

‖

3^{3}

=

27

\frac{6}{6} \cdot x^{3}

+ \frac{18}{2} \cdot x^{2}

+ \frac{54}{2} \cdot x

+ \frac{6}{6} \cdot 27

ℹ

\frac{6}{6} \cdot x^{3}

=

\frac{6 x^{3}}{6}

‖

\frac{6 x^{3}}{6}

=

x^{3}

‖

\frac{18}{2} \cdot x^{2}

=

\frac{18 x^{2}}{2}

‖

\frac{18 x^{2}}{2}

=

9 x^{2}

‖

\frac{54}{2} \cdot x

=

\frac{54 x}{2}

‖

\frac{54 x}{2}

=

27 x

‖

\frac{6}{6} \cdot 27

=

\frac{6 \cdot 9}{2}

x^{3}

+ 9 x^{2}

+ 27 x

+ \frac{6 \cdot 9}{2}

ℹ

6 \cdot 9

=

54

x^{3}

+ 9 x^{2}

+ 27 x

+ \frac{54}{2}

▷Résultat : $x^{3}$ $+ 9 x^{2}$ $+ 27 x$ $+ 27$