To cube a binomial we can first rewrite it as the product of a first and a second power.

(a + b)^{3} = (a + b) (a + b)^{2}

We can then expand the squared binomial, obtaining a perfect square trinomial. Finally we multiply the linear binomial left over with the obtained trinomial.

(a + b)^{3}

Rewrite

Rewrite $(a + b)^{3}$ as $(a + b) (a + b)^{2}$

(a + b) (a + b)^{2}

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

(a + b) (a^{2} + 2 a b + b^{2})

Distr

Distribute $(a^{2} + 2 a b + b^{2})$

a (a^{2} + 2 a b + b^{2}) + b (a^{2} + 2 a b + b^{2})

Multiply

a^{3} + 2 a^{2} b + a b^{2} + a^{2} b + 2 a b^{2} + b^{3}

AddTerms

Add terms

a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

As we can see, this procedure is is not very efficient. However, we can identify a specific pattern in our result.

a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} ⇕ (1) a^{3} b^{0} + 3 a^{2} b^{1} + 3 a^{1} b^{2} + (1) a^{0} b^{3}

The exponents for the binomial's first term are decreasing by one, starting from

3

until reaching

0 .

Those for the other term are increasing by one, starting from

0

until reaching

3 .

Furthermore, the coefficients of the resulting terms coincide with the numbers from row

3

of Pascal's triangle.

Row 0123 Pascal ’ s Triangle 1113121311

We can use this pattern to cube a binomial in a more convenient way. Let's see an example. We will calculate

(x + 2)^{3}

by following this pattern.

(x + 2)^{3} (x + 2)^{3} (x + 2)^{3} = (1) x^{3} (2)^{0} + 3 x^{2} (2)^{1} + 3 x^{1} (2)^{2} + (1) x^{0} (2)^{3} = (1) x^{3} (1)^{2} + 3 x^{2} (2)^{2} + 3 x (4)^{2}^{2} + (1) (1) (8) = x^{3} + 6 x^{2} + 12 x + 8

This is faster and easier than following the procedure that we did at first for a general case

(a + b)^{3} .

Exercise