# 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. (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).