# Cube of Binomial

In this chapter we will learn the concept & formula for cube of binomial. In the end we will also see the solved problems related to the concept.

Let us first understand the basics of binomial.

## What is Binomial?

Binomial is an algebraic expression with two entities.

The entities can be variable, constant or both.

Examples of binomial are given below;

\mathtt{\Longrightarrow \ 6x\ +\ 5y}\\\ \\ \mathtt{\Longrightarrow \ 2\ +\ 3x}\\\ \\ \mathtt{\Longrightarrow \ 8x^{3} +9xy}\\\ \\ \mathtt{\Longrightarrow \ 12xy^{2} +x^{2} y}

## Cube of Binomial formula

There are two important formula for cube of binomial;

(i) \mathtt{ \ \ ( a+b)^{3} =a^{3} +b^{3} +3ab\ ( a+b)}\\\ \\ \mathtt{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }

(ii) \mathtt{\ ( a-b)^{3} =a^{3} -b^{3} -3ab\ ( a-b)}\\\ \\ \mathtt{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }

You need to memorize both the formulas as they are important to solved cube of binomials problems.

## Cube of Binomial solved problem

Example 01
Find the cube of binomial 10x – 3y

Solution
The cube of given binomial is expressed as \mathtt{( 10x-3y)^{3}}

Here we will use the below formula;

\mathtt{( a-b)^{3} =a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }

Here;
a = 10x
b = 3y

Putting the values;

\mathtt{\Longrightarrow \ ( 10x-3y)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 10x)^{3} -( 3y)^{3} -3( 10x)^{2}( 3y) +3( 10x)( 3y)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 1000x^{3} -27y^{3} -900x^{2} y+270xy^{2}}

The above expression is the cube of given binomial.

Example 02
Find the cube of binomial 5x + 2y

Solution
The cube of binomial is expressed as \mathtt{( 5x+2y)^{3}}

We will use the formula;
\mathtt{( a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }

Putting the values;
\mathtt{\Longrightarrow \ ( 5x+2y)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x)^{3} +( 2y)^{3} +3( 5x)^{2}( 2y) +3( 5x)( 2y)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 125x^{3} +8y^{3} +150x^{2} y+60xy^{2}}

The above expression is the cube of given binomial.

Example 03
Expand \mathtt{( x+2y)^{3} +( x-2y)^{3}}

Solution
The questions involve cube of two binomial.

We will solve the question in two steps.

First expand \mathtt{( x+2y)^{3}}

\mathtt{\Longrightarrow \ ( x+2y)^{3}}\\\ \\ \mathtt{\Longrightarrow ( x)^{3} +( 2y)^{3} +3( x)^{2}( 2y) +3( x)( 2y)^{2}}\\\ \\ \mathtt{\Longrightarrow x^{3} +8y^{3} +6x^{2} y+12xy^{2}}

Now expand \mathtt{( x-2y)^{3}}

\mathtt{\Longrightarrow \ ( x-2y)^{3}}\\\ \\ \mathtt{\Longrightarrow ( x)^{3} -( 2y)^{3} -3( x)^{2}( 2y) +3( x)( 2y)^{2}}\\\ \\ \mathtt{\Longrightarrow x^{3} -8y^{3} -6x^{2} y+12xy^{2}}

\mathtt{\Longrightarrow \ ( x+2y)^{3} +( x-2y)^{3}}\\\ \\ \mathtt{\Longrightarrow x^{3} +\cancel{8y^{3}} +\cancel{6x^{2} y} +12xy^{2} +x^{3} -\cancel{8y^{3}} -\cancel{6x^{2} y} +12xy^{2}}\\\ \\ \mathtt{\Longrightarrow \ 12x^{3} +24xy^{2}}

Hence, the above expression is solution of given binomials.

Example 04
Find the value of \mathtt{( 19)^{3}} using cube of binomial formula.

Solution
The number can be converted into binomial as;

\mathtt{\Longrightarrow \ ( 19)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 20-1)^{3}}

We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }

Putting the values;

\mathtt{\Longrightarrow \ ( 20-1)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( 20)^{3} -( 1)^{3} -3( 20)^{2}( 1) +3( 20)( 1)^{2}}\\\ \\ \mathtt{\Longrightarrow \ 8000-1-1200\ +60}\\\ \\ \mathtt{\Longrightarrow \ 6859}

Hence, 6859 is the value of given number.

Example 05
Find the coefficient of \mathtt{x^{2} y} in the expression \mathtt{( 6x+y)^{3}}

Solution
We will use the formula;
\mathtt{( a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }

Putting the values;

\mathtt{\Longrightarrow \ ( 6x+y)^{3}}\\\ \\ \mathtt{\Longrightarrow ( 6x)^{3} +( y)^{3} +3( 6x)^{2}( y) +3( 6x)( y)^{2}}\\\ \\ \mathtt{\Longrightarrow 216x^{3} +y^{3} +78x^{2} y\ +18xy^{2}}

Here the coefficient of \mathtt{x^{2} y} is 78

Example 06
Expand \mathtt{\left(\frac{1}{2} x+y\right)^{3} -\left(\frac{1}{2} x-y\right)^{3}}

Solution
The question involve cube of two binomial.

We will solve the question in two steps;

First expand \mathtt{\left(\frac{1}{2} x+y\right)}

Using the formula;
\mathtt{( a+b)^{3} =a^{3} +b^{3} +3a^{2} b+3ab^{2} \ }

Put the values;
\mathtt{\Longrightarrow \ \left(\frac{1}{2} x+y\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \left(\frac{1}{2} x\right)^{3} +( y)^{3} +3\left(\frac{1}{2} x\right)^{2}( y) +3\left(\frac{1}{2} x\right)( y)^{2}}\\\ \\ \mathtt{\Longrightarrow \frac{\mathtt{x^{3}}}{8} +y^{3} +\frac{3}{4} x^{2} y\ +\frac{3}{2} xy^{2}}

Now expand \mathtt{\left(\frac{1}{2} x-y\right)^{3}}

We will use the formula;
\mathtt{( a-b)^{3} =a^{3} -b^{3} -3a^{2} b+3ab^{2} \ }

Putting the values;

\mathtt{\Longrightarrow \ \left(\frac{1}{2} x-y\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \left(\frac{1}{2} x\right)^{3} -( y)^{3} -3\left(\frac{1}{2} x\right)^{2}( y) +3\left(\frac{1}{2} x\right)( y)^{2}}\\\ \\ \mathtt{\Longrightarrow \frac{\mathtt{x^{3}}}{8} -y^{3} -\frac{3}{4} x^{2} y\ +\frac{3}{2} xy^{2}}

Joining both the parts;

\mathtt{\Longrightarrow \ \left(\frac{1}{2} x+y\right)^{3} -\left(\frac{1}{2} x-y\right)^{3}}\\\ \\ \mathtt{\Longrightarrow \cancel{\mathtt{\frac{\mathtt{x^{3}}}{8}}} +y^{3} +\frac{3}{4} x^{2} y\ +\cancel{\mathtt{\frac{3}{2} xy^{2}}} -\cancel{\mathtt{\frac{\mathtt{x^{3}}}{8}}} +y^{3} +\frac{3}{4} x^{2} y\ -\cancel{\mathtt{\frac{3}{2} xy^{2}}}}\\\ \\ \mathtt{\Longrightarrow \ 2y^{3} +\frac{6}{4} x^{2} y}

Hence, above expression is solution of given binomial