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}}