In this mini-lesson, we will learn about the cube of a binomial by understanding methods to find the cube of a binomial, and how to apply them while solving problems. We will also discover interesting facts around them.

John built a cubical water tank.

When his friend asked him what the length of the cube's side was. John replied it was \(a + b\).

His friend then asked him how much water it could hold. John then suggested they cube the binomial, as the volume of a cube is \(l^3\).

So, that volume of the cubical tank can be obtained by

\[{(a + b)}^3 = a^3 + b^3 +3ab(a + b)\]

The above equation is a cube of the binomial.

To write the cube of any number we simply multiply it three times to itself.

For example, cube of \(2\) is \(2^3 = 2 \times 2 \times 2 = 8\)

Similarly, the cube of a binomial \((a + b)\) is

\[{(a + b)}^3 = (a + b) \times (a + b) \times (a + b)\]

Let's explore the cube of a binomial in more detail

**Lesson Plan**

**How to Find the Cube of a Binomial?**

Let's consider a basic binomial \({(x + y)}\).

Cube of the above binomial is

\({(x + y)}^3 = x^3 + y^3 + 3xy(x + y)\) |

The above result is also an identity it holds true for every value of \(x\) and \(y\).

If we replace \(y\) with \(-y\) we can get.

\({(x - y)}^3 = x^3 - y^3 - 3xy(x - y)\) |

We can see that every \(+\) sign in the expansion of \({(x + y)}^3\) has changed to a \(-\) sign in the expansion of \({(x - y)}^3\)

**Cubing a Binomial Steps**

Let's see the steps to obtaining the above result for the binomial \({(x + y)}\).

**Step 1: **First write the cube of the binomial \({(x + y)}^3 = (x + y) \times (x + y) \times (x + y)\)

**Step 2: **Multiply the first two binomials and keep the third one as it is

\[\begin{align} {(x + y)}^3 &= (x + y) \times (x + y) \times (x + y) \\[0.2cm]

{(x + y)}^3 &= [x(x + y) + y(x + y)](x + y) \\[0.2cm]

{(x + y)}^3 &= [x^2 + xy + xy + y^2](x + y) \\[0.2cm]

{(x + y)}^3 &= [x^2 + 2xy + y^2](x + y) \end{align}\]

**Step 3: **Multiply the remaining binomial to the trinomial so obtained

\[\begin{align} {(x + y)}^3 &= [x^2 + 2xy + y^2](x + y) \\[0.2cm]

{(x + y)}^3 &= x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2) \\[0.2cm]

{(x + y)}^3 &= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3 \\[0.2cm]

{(x + y)}^3 &= x^3 + 2x^2y + x^2y + xy^2 + 2xy^2 + y^3 \\[0.2cm]

{(x + y)}^3 &= x^3 + 3x^2y + 3xy^2 + y^3 \\[0.2cm]

{(x + y)}^3 &= x^3 + y^3 + 3x^2y + 3xy^2 \\[0.2cm]

{(x + y)}^3 &= x^3 + y^3 + 3xy(x + y) \end{align}\]

**Examples**

Let's take a binomial in one variable \((x + 1)\).

The cube of the above binomial can be obtained following the above steps.

**Step 1: **First write the cube of the binomial \({(x + 1)}^3 = (x + 1) \times (x + 1) \times (x + 1)\)

**Step 2: **Multiply the first two binomials and keep the third one as it is.

\[\begin{align} {(x + 1)}^3 &= (x + 1) \times (x + 1) \times (x + 1) \\[0.2cm]

{(x + 1)}^3 &= [x(x + 1) + 1(x + 1)](x + 1) \\[0.2cm]

{(x + 1)}^3 &= [x^2 + x \times 1 + x \times 1 + 1^2](x + 1) \\[0.2cm]

{(x + 1)}^3 &= [x^2 + 2x + 1](x + 1) \end{align}\]

**Step 3: **Multiply the remaining binomial to the trinomial so obtained.

\[\begin{align} {(x + 1)}^3 &= [x^2 + 2x + 1](x + 1) \\[0.2cm]

{(x + 1)}^3 &= x(x^2 + 2x + 1) + 1(x^2 + 2x + 1) \\[0.2cm]

{(x + 1)}^3 &= x^3 + 2x^2 + x + x^2 + 2x + 1 \\[0.2cm]

{(x + 1)}^3 &= x^3 + 3x^2 + 3x + 1 \\[0.2cm]

{(x + 1)}^3 &= x^3 + 1 + 3x(x + 1) \end{align}\]

Cube of the binomial \((x - 1)\) can now be written as

\[{(x - 1)}^3 = x^3 - 1 - 3x(x - 1) \]

Explore the cube of a binomial calculator below to learn more about the steps of cubing a binomial.

- Binomials are polynomials with only two terms.
- Identity in algebra is the mathematical equation that holds true for every value of the variable.
- The cube of a negative number or variable is also negative.

**Solved Examples**

Example 1 |

Kevin cubed a binomial \(2x + 3y\) and finds out that the coefficient of \(x^y\) is 36, is he right?

**Solution**

Cube of the binomial \(2x + 3y\) is \({(2x + 3y)}^3\)

We can use the identity \({(a + b)}^3 = a^3 + b^3 + 3ab(a + b)\)

Replace \(a\) with \(2x\) and \(b\) with \(2y\)

\[\begin{align} {(2x + 3y)}^3 &= {(2x)}^3 + {(3y)}^3 + 3(2x)(3y)(2x + 3y) \\[0.2cm]

{(2x + 3y)}^3 &= 8x^3 + 27y^3 + 18xy(2x + 3y) \\[0.2cm]

{(2x + 3y)}^3 &= 8x^3 + 27y^3 + 18xy \times 2x + 18xy \times 3y \\[0.2cm]

{(2x + 3y)}^3 &= 8x^3 + 27y^3 + 36x^y + 54xy^2 \end{align}\]

Coefficient of \(x^2y\) is 36

\(\therefore\) Kevin is right |

Example 2 |

Mathew told to his friend that he can find the cube of the number \(999\) without even multiplying it three times, explain how he finds the cube of \(999\)?

**Solution**

The number \(999\) can be written as \(1000 - 1\)

\({(999)}^3 = {(1000 - 1)}^3\)

Using the identity \({(a - b)}^3 = a^3 - b^3 - 3ab(a - b)\)

Replace \(a\) with \(1000\) and \(b\) with \(1\)

\[\begin{align} {(1000 - 1)}^3 &= {(1000)}^3 - 1^3 - 3(1000)(1)(1000 - 1) \\[0.2cm]

{(1000 - 1)}^3 &= 1000000000 - 1 - 3(1000)(1)(999) \\[0.2cm]

{(1000 - 1)}^3 &= 1000000000 - 1 - 2997000 \\[0.2cm]

{(1000 - 1)}^3 &= 997002999 \end{align}\]

\(\therefore\) \({(999)}^3 = 997002999\) |

- Find \(a\) if the coefficients of \(x^2\) and \(x^3\) in the expansion of \({(3+ax)}^9\) are equal.

**Interactive Questions on Cube of a Binomial**

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**Let's Summarize**

The mini-lesson targeted in the fascinating concept of the cube of a binomial. The math journey around the cube of binomial starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

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**FAQs on Cube of a Binomial**

### 1. How do you simplify the cube of a binomial?

Cube of a binomial can be simplified using the identities:

\({(x + y)}^3 = x^3 + y^3 - 3xy(x + y)\)

\({(x - y)}^3 = x^3 - y^3 - 3xy(x - y)\)

### 2. What is a binomial?

A binomial is a type of polynomial that only has two terms.

For example, \(x + 2y\), \(a^3 + 2b\).

### 3. How do you expand a binomial?

A binomial can be expanded using the identities:

\({(x + y)}^3 = x^3 + y^3 - 3xy(x + y)\)

\({(x - y)}^3 = x^3 - y^3 - 3xy(x - y)\)

### 4. How do you expand a binomial using Pascal’s Triangle?

We can expand binomials using Pascal's triangle like:

### 5. How do you cube expressions?

An expression can be cubed by multiplying itself three times.

For example, \({(a + b)}^3 = (a + b)(a + b)(a + b)\)

### 6. What are the steps for cubing a binomial?

**Step 1: **First write the cube of the binomial \({(x + y)}^3 = (x + y) \times (x + y) \times (x + y)\).

**Step 2: **Multiply the first two binomials and keep the third one as it is.

**Step 3: **Multiply the remaining binomial to the trinomial so obtained.

### 7. What do you mean by the cube of a binomial?

A cube of a binomial is multiplying the binomial three times to itself.

### 8. How do you find the cube of a binomial?

A cube of a binomial can be found by multiplying to itself three times.

Or we can find the cube by using identity

\({(x + y)}^3 = x^3 + y^3 - 3xy(x + y)\)

\({(x - y)}^3 = x^3 - y^3 - 3xy(x - y)\)