\$\$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. 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^2lue(a-b) + ablue(a-b) + b^2lue(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^2lue(x - 3) + 3xlue(x - 3) + 9lue(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* \$\$