# Factorize algebraic expressions with difference of cubes

In this chapter, we will learn to factorize algebraic expressions containing difference of cubes with the help of a formula.

At the end of the chapter, some solved problems are also given for practice purpose.

## How to factorize difference of cubes?

To factorize the difference of cubes expression, you have to use following formula;

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

This formula is very important.

You have to memorize the formula for solving these types of expression.

I hope you understood the concept. Let us solve some problems for practice purpose.

## Factorizing difference of cubes – Solved problems

Example 01
Solve the below expression.

\mathtt{\Longrightarrow \ x^{3} -125}

Solution
The above expression can be written as;

\mathtt{\Longrightarrow \ x^{3} -5^{3}}

Referring the formula;
\mathtt{a^{3} -b^{3} =\ ( a-b)\left( a^{2} +ab+b^{2}\right)}

Where;
a = x
b = 5

Putting the values in formula, we get;

\mathtt{\Longrightarrow \ ( x\ -\ 5)\left( x^{2} +5x\ +\ 5^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x\ -\ 5)\left( x^{2} +5x\ +\ 25\right)}

Hence, the above expression is the factorized form.

Example 02
Factorize the below expression.

\mathtt{\Longrightarrow \ 1331x^{3} -8y^{3}}

Solution
The above expression can be written as;

\mathtt{\Longrightarrow \ ( 11x)^{3} -( 2y)^{3}}

Referring the formula;
\mathtt{a^{3} -b^{3} =\ ( a-b)\left( a^{2} +ab+b^{2}\right)}

Where;
a = 11x
b = 2y

Putting the values in formula, we get;

\mathtt{\Longrightarrow \ ( 11x\ -\ 2y)\left(( 11x)^{2} +11x.2y\ +\ ( 2y)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( 11x\ -\ 2y)\left( 121x^{2} +22xy\ +\ 4y^{2}\right)}

Hence, the above expression is the solution.

Example 03
Factorize the below expression

\mathtt{\Longrightarrow \ 3y^{3} -24}

Solution
The above expression can be written as;

\mathtt{\Longrightarrow \ 3\ \left( y^{3} -8\right)}\\\ \\ \mathtt{\Longrightarrow \ 3\ \left( y^{3} -2^{3}\right)}

Referring the formula;
\mathtt{a^{3} -b^{3} =\ ( a-b)\left( a^{2} +ab+b^{2}\right)}

Where;
a = y
b = 2

Using the formula, we get;

\mathtt{\Longrightarrow \ 3\ ( y-2)\left( y^{2} +2y+2^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 3\ ( y-2)\left( y^{2} +2y+4\right)}

Hence, the above expression is the solution.

Example 04
Factorize the below expression;
\mathtt{\Longrightarrow \ 27x^{3} y^{9} -343}

Solution
The above expression can be written as;

\mathtt{\Longrightarrow \ 3^{3} x^{3}\left( y^{3}\right)^{3} -7^{3}}\\\ \\ \mathtt{\Longrightarrow \ \left( 3xy^{3}\right)^{3} -7^{3}}

Referring the formula;
\mathtt{a^{3} -b^{3} =\ ( a-b)\left( a^{2} +ab+b^{2}\right)}

Where;
a = \mathtt{3xy^{3}}
b = 7

Putting the values we get;

\mathtt{\Longrightarrow \ \left( 3xy^{3} -7\right)\left(\left( 3xy^{3}\right)^{2} +3xy^{3} .7+7^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( 3xy^{3} -7\right)\left( 9x^{2} y^{6} +21xy^{3} +49\ \right)}

Hence, the above expression is the solution.

Next chapter : Factorizing expression \mathtt{a^{3} +b^{3}}

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