The difference of two cube comes under algebra, which is one of the important sub-part of mathematics. Algebra is the study of mathematical symbols, functions, equations, etc. Factoring polynomial is a operation of numbers. By using factoring polynomial or equations it is easy to solve them. Using Difference of two cubes formula we can factor complex polynomials also. Difference of two cube is one of the formula of algebra. In this page we are going to discuss about all properties of Difference of two cubes. |

It is an algebraical method of factorizing polynomial. The difference of cube is define as,

"The product of two numbers subtraction and the square root of two number's sum".

"The product of two numbers subtraction and the square root of two number's sum".

Let two number are a and b then the difference of two cube is denoted by $a^{3}-b^{3}$.

This method can be use for two perfect cubes. Difference of two cube is also used for complex numbers.

Let we have two number a and b which are perfect cube, then he formula is,

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

**Proof:**

RHS $(a-b)(a^{2}+ab+b^{2})$

$a^{3} + a^{2}b + ab^{2} - ba^{2} - ab^{2} - b^{3}$

$a^{2}b = ba^{2} = 0$

$ab^{2} = b^{2}a = 0$

Hence we left with ,

$a^{3}-b^{3}$

There for it is proved

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

As we have already discussed about the difference of two cubes above. So the sum of The Sum of two cubes is the product of a trinomial and a binomial. Let two number are a and b then the sum of two cubes is,

**Proof:**

RHS $(a+b)(a^{2}-ab+b^{2})$

$a^{3} - a^{2}b + ab^{2} + ba^{2} - ab^{2} + b^{3}$

$a^{2}b = ba^{2} = 0$

$ab^{2} = b^{2}a = 0$

Hence we left with ,

$a^{3} + b^{3}$

There for it is prooved

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

Factoring the difference of two cubes is divided into 4 steps. Follow the below given steps;

**Step 1:**Find if there is any common factor given in the polynomial.

**Step 2:**Take out the common factor of the polynomial.

**Step 3:**Using above given formula find the value of a and b

**Step 4:**Simplify the factorize the polynomial.

**Example:**

Find the cube of the polynomial $(27^{3} - 9^{3})$

**Solution:**$(27^{3} - 9^{3})$

**Step 1:**Here the common factor is 3

= $3 (9^{3} - 3^{3})$

Form given a = 9 and b = 3

**Step 2:**Applying the formula $(a^{3}-b^{3}) = (a-b)(a^{2}+ab+b^{2})$

We have,

= $3(9^{3}-3^{3})$

= $3(9-3)(9^{2}+9*3+3^{2})$

= $3(6)(81+27+9)$

=$3(6)(117)$

=$3*702$

=$2106$

**Example 1:**

Find the factorization of the polynomial $a^{3} - 216$

**Solution:**

Given $a^{3} - 216$

Here $a^{3}$ is the perfect cube.

And 216 is the perfect cube of 6.

Now we have;

$a^{3} - 6^{3}$

From the given we have

a = a and b = 6

Applying formula;

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

$(a^{3}-6^{3}) = (a-6)(a^{2}+a6+6^{2})$

$(a^{3}-6^{3}) = (a-6)(a^{2}+6a+36)$

Hence the factorization of $(a^{3}-6^{3})$ is $(a-6)(a^{2}+6a+36)$

**Example 2:**

Find the factorization of the polynomial $m^{3} - 512$

**Solution:**

Given $m^{3} - 512$

Here $x^{3}$ is the perfect cube.

And 512 is the perfect cube of 8.

Now we have;

$x^{3} - 8^{3}$

From the given we have

a = x and b = 8

Applying formula;

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

$(x^{3}-8^{3}) = (x-8)(x^{2}+8x+8^{2})$

$(x^{3}-8^{3}) = (x-8)(x^{2}+8x+64)$

Hence the factorization of $(x^{3}-8^{3})$ is $(x-8)(x^{2}+8x+64)$