expand (x-1)^3 — step-by-step solution | Explain MAths

expand (x-1)^3

Let's expand:

{(x - 1)}^{3}

With Newton's binomial formula:

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

, let's expand the expression:

\frac{3!}{3! \cdot 0!} \cdot x^{3} \cdot {(-1)}^{0}

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

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

+ \frac{3!}{0! \cdot 3!} \cdot x^{0} \cdot {(-1)}^{3}

ℹ

{(-1)}^{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

‖

{(-1)}^{1}

=

-1

‖

\frac{6}{2} \cdot x^{2} \cdot -1

=

- \frac{6}{2} \cdot x^{2}

‖

3!

=

6

‖

1!

=

1

‖

2!

=

2

‖

1 \cdot 2

=

2

‖

x^{1}

=

x

‖

{(-1)}^{2}

=

1

‖

\frac{6}{2} \cdot x \cdot 1

=

\frac{6}{2} \cdot x

‖

x^{0}

=

1

‖

\frac{3!}{0! \cdot 3!} \cdot 1 \cdot {(-1)}^{3}

=

\frac{3!}{0! \cdot 3!} \cdot {(-1)}^{3}

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

- \frac{6}{2} \cdot x^{2}

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

+ \frac{3!}{0! \cdot 3!} \cdot {(-1)}^{3}

ℹ

3!

=

6

‖

3!

=

6

‖

0!

=

1

‖

6 \cdot 1

=

6

‖

- \frac{6}{2} \cdot x^{2}

=

- \frac{6 x^{2}}{2}

‖

- \frac{6 x^{2}}{2}

=

-3 x^{2}

‖

\frac{6}{2} \cdot x

=

\frac{6 x}{2}

‖

\frac{6 x}{2}

=

3 x

‖

3!

=

6

‖

0!

=

1

‖

3!

=

6

‖

1 \cdot 6

=

6

‖

{(-1)}^{3}

=

-1

‖

\frac{6}{6} \cdot -1

=

- \frac{6}{6}

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

- 3 x^{2}

+ 3 x

- \frac{6}{6}

▷Answer: $x^{3}$ $- 3 x^{2}$ $+ 3 x$ $- 1$