Sum and Difference of Cubes

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Sum and Difference of Cubes:

$$$\huge{\color{purple}{a^3 \pm b^3=\left(a\pm b\right)\left(a^2 \mp ab+b^2 \right)}}$$$

Proof of this fact is straightforward.

We just prove it from right to left.

Multiply polynomials:

`(a-b)(a^2+ab+b^2)=a*a^2+a*ab+a*b^2-b*a^2-b*ab-b*b^2=`

`=a^3+a^2b+ab^2-a^2b-ab^2-b^3=a^3-b^3`.

Similarly, it can be shown, that `a^3+b^3=(a+b)(a^2-ab+b^2)`.

Or written more compactly: $$$a^3 \pm b^3=\left(a\pm b\right)\left(a^2 \mp ab+b^2 \right)$$$

Expressions `a^2+ab+b^2` and `a^2-ab+b^2` are often called incomplete squares, because they lack one `ab` to become perfect square (`(a+-b)^2=a^2+-color(red)(2)ab+b^2`).

Don't attempt to factor incomplete square. It can't be factored.

Example 1. Factor `x^3+8`.

Notice, that `8=2^3`.

Thus, `x^3+8=x^3+2^3=(x+2)(x^2-x*2+2^2)=(x+2)(x^2-2x+4)`.

Answer: `x^3+8=(x+2)(x^2-2x+4)`.

Of course, there can be more complex expressions.

Example 2. Factor `27y^3-64`.

Notice, that `27y^3=(3y)^3` and `64=4^3`.

Thus, `27y^3-64=(3y)^3-4^3=(3y-4)((3y)^2+3y*4+4^2)=(3y-4)(9y^2+12y+16)`.

Answer: `27y^3-64=(3y-4)(9y^2+12y+16)`.

And even harder...

Example 3. Factor the following: `8m^6n^9+27a^9b^3`.

Notice, that `8m^6n^9=(2m^2n^3)^3` and `27a^9b^3=(3a^3b)^3`.

Thus, `8m^6n^9+27a^9b^3=(2m^2n^3)^3+(3a^3b)^3=`

`=(2m^2n^3+3a^3b)((2m^2n^3)^2-2m^2n^3*3a^3b+(3a^3b)^2)=`

`=(2m^2n^3+3a^3b)(4m^4n^6-6a^3bm^2n^3+9a^6b^2)`.

Answer: `8m^6n^9+27a^9b^3=(2m^2n^3+3a^3b)(4m^4n^6-6a^3bm^2n^3+9a^6b^2)`.

Now, it is time to exercise.

Exercise 1. Factor the following: `n^3+125`.

Answer: `(n-5)(n^2+5n+25)`.

Exercise 2. Factor the following: `-343+y^6x^3`.

Answer: `(y^2x-7)(y^4x^2+7y^2x+49)`. Hint: `-343+y^6x^3=y^6x^3-343`.

Exercise 3. Factor `x^3y^3z^3-27a^3`.

Answer: `(xyz-3a)(x^2y^2z^2+3axyz+9a^2)`.

Exercise 4. Factor `8x^3+1`.

Answer: `(2x+1)(4x^2-2x+1)`.