$$a^3 - b^3 = (a - b)(a^2 + abdominal muscle + b^2)$$ The discussion listed below includes 3 separate methods of demonstrating that this formula is valid.
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on the page around the amount of cubes we proved that $$a^3 + b^3 = a^2 - abdominal muscle + b^2$$. We have the right to use this formula to find a factorization for $$a^3 - b^3$$.

We begin by writing $$a^3 - b^3$$ as $$a^3 + (-b)^3$$ and also then using the sum of cubes pattern.

$$ eginalign* a^3 - b^3 & = a^3 + (-b)^3\ & = left(a + (-b) ight)left(a^2 + a(-b) + (-b)^2 ight)\ & = (a - b)(a^2 - ab + b^2) endalign* $$

we could also determine the validity of the formula using the same techniques we developed in the amount of cubes lesson.


Explanation of the Formula -- straight Method


We have the right to verify the factoring formula by expanding the an outcome and seeing the it simplifies come the original, together follows.

$$ eginalign* lue(a-b)(a^2 + abdominal muscle + b^2) & = a^2lue(a-b) + ablue(a-b) + b^2lue(a-b)\ & = a^3 - a^2b + a^2b - ab^2 + ab^2 - b^3\ & = a^3 lue- a^2b + a^2b,, ed - ab^2 + ab^2 - b^3\ & = a^3 + lue 0 + ed 0 - b^3\ & = a^3 - b^3 endalign* $$


Explanation that the Formula---Division Method


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

action 1

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

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

one of the solutions is $$a = b$$. Adding $$b$$ to both political parties of this equation offers us $$a - b = 0$$, which means $$(a - b)$$ is a aspect of $$a^3 - b^3$$.

step 2

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

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

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


present that $$x^3 - 27$$ factors into $$(x - 3)(x^2 + 3x + 9)$$.

step 1

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

$$ eginalign* lue(x - 3)(x^2 + 3x + 9) & = x^2lue(x - 3) + 3xlue(x - 3) + 9lue(x - 3)\ & = x^3 - 3x^2 + 3x^2 - 9x + 9x - 27\ & = x^3 - 27 endalign* $$

step 1 (Alternate Solution)

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

Since we want to variable $$x^3 - 27$$, we very first identify $$a$$ and also $$b$$.

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because $$a$$ is the cube root of the an initial term, we recognize $$a = sqrt<3>x^3 = x$$.

Likewise, because $$b$$ is the cube root of the 2nd term, we know $$b = sqrt<3>27 = 3$$.

step 2

Write under the factored form.

$$ eginalign* a^3 + b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2)\ x^3 - 27 & = (lue x - ed 3)(lue x^2 + lue xcdot ed 3 + ed 3^2)\ & = (x-3)(x^2 + 3x + 9) endalign* $$