$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ The discussion below includes 3 separate methods of demonstrating that this formula is valid.

on the page around the amount of cubes we confirmed that $$a^3 + b^3 = a^2 - abdominal muscle + b^2$$. We have the right to use this formula to find a factorization because that $$a^3 - b^3$$.

We begin by creating $$a^3 - b^3$$ as $$a^3 + (-b)^3$$ and then utilizing the amount of cubes pattern.

\beginalign* a^3 - b^3 & = a^3 + (-b)^3\\ & = \left(a + (-b)\right)\left(a^2 + a(-b) + (-b)^2\right)\\ & = (a - b)(a^2 - ab + b^2) \endalign*

us could also determine the validity that the formula utilizing the same approaches we developed in the amount of cubes lesson.

### Explanation the the Formula -- straight Method

We deserve to verify the factoring formula by widening the result and seeing the it simplifies come the original, as follows.

\beginalign* \blue(a-b)(a^2 + abdominal + b^2) & = a^2\blue(a-b) + ab\blue(a-b) + b^2\blue(a-b)\\ & = a^3 - a^2b + a^2b - ab^2 + ab^2 - b^3\\ & = a^3 \blue- a^2b + a^2b\,\,\red - ab^2 + ab^2 - b^3\\ & = a^3 + \blue 0 + \red 0 - b^3\\ & = a^3 - b^3 \endalign*

### Explanation of the Formula---Division Method

Another way to confirm to the formula is to find a systems to $$a^3 - b^3 = 0$$, and then use department to uncover the factored form.

action 1

Find a solution to $$a^3 - b^3 = 0$$.

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\beginalign* a^3 - b^3 & = 0\\ a^3 & = b^3\\ \sqrt<3>a^3 & = \sqrt<3>b^3\\ a & = b \endalign*

one of the solutions is $$a = b$$. Adding $$b$$ to both sides of this equation provides us $$a - b = 0$$, which way $$(a - b)$$ is a aspect of $$a^3 - b^3$$.

step 2

Find the other variable of $$a^3 - b^3$$ utilizing polynomial division.

\beginalign* \beginarrayrrcrcrcr & a^2 \hspace2mm +ab \hspace2mm +b^2 \hspace22mm \\ & a - b\,\, \encloselongdiva^3 + 0\, a^2b + 0\, ab^2 - b^3 \hspace10mm \\ & \underline-(a^3 - a^2b) \hspace35mm \\ & a^2b + 0\,ab^2 - b^3\ \hspace12mm \\ & - \underline(a^2b - ab^2) \hspace23mm\\ & ab^2 - b^3 \hspace15mm \\ & - \underlineab^2 - b^3) \hspace13mm\\ & 0 \hspace15mm \endarray \endalign*

Since $$a - b$$ divides evenly right into $$a^3 - b^3$$, us know

$$(a - b)(a^2 + abdominal muscle + b^2) = a^3 - b^3$$

### distinction Of Cubes Calculator

show that $$x^3 - 27$$ determinants into $$(x - 3)(x^2 + 3x + 9)$$.

step 1

present that broadening $$(x - 3)(x^2 + 3x + 9)$$ outcomes in $$x^3 - 27$$.

\beginalign* \blue(x - 3)(x^2 + 3x + 9) & = x^2\blue(x - 3) + 3x\blue(x - 3) + 9\blue(x - 3)\\ & = x^3 - 3x^2 + 3x^2 - 9x + 9x - 27\\ & = x^3 - 27 \endalign*

action 1 (Alternate Solution)

show that $$(x - 3)(x^2 + 3x + 9)$$ matches the exactly pattern for the formula.

Since we want to aspect $$x^3 - 27$$, we an initial identify $$a$$ and also $$b$$.

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because $$a$$ is the cube root of the first term, we know $$a = \sqrt<3>x^3 = x$$.

Likewise, since $$b$$ is the cube source of the 2nd term, we understand $$b = \sqrt<3>27 = 3$$.

step 2

Write down the factored form.

\beginalign* a^3 + b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ x^3 - 27 & = (\blue x - \red 3)(\blue x^2 + \blue x\cdot \red 3 + \red 3^2)\\ & = (x-3)(x^2 + 3x + 9) \endalign*