Thread: sum and differences of cubes

how would you solve these by factoring?

#1. (x+2)^3 - (x-2)^3
#2. 16(x+1)^3 + 2
#3. (x+1)^3 - (y-2)^3

thanks

Originally Posted by checkmarks

how would you solve these by factoring?

#1. (x+2)^3 - (x-2)^3
#2. 16(x+1)^3 + 2
#3. (x+1)^3 - (y-2)^3

thanks

For starters, you aren't "solving" your are "reducing."

Recall that the sum and difference of two perfect cubes factors:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

1. (x + 2)^3 - (x - 2)^3
Here we have the difference of two cubes: let a = x + 2 and b = x - 2. Then
(x + 2)^3 - (x - 2)^3 = [(x + 2) - (x - 2)][(x + 2)^2 + (x + 2)(x - 2) + (x - 2)^2]

= (4)[(x^2 + 4x + 4) + (x^2 - 4) + (x^2 - 4x + 4)] = 4(3x^2 + 4)

2. 16(x + 1)^3 + 2 = 2[8(x + 1)^3 + 1]
Here we have the sum of two cubes: a = 2(x + 1) and b = 1:
16(x + 1)^3 + 2 = 2[8(x + 1)^3 + 1] = 2[{2(x + 1)} + {1}][{2(x + 1)}^2 - {2(x + 1)}{1} + {1}^2}

= 2[2x + 3][{2(x^2 + 2x + 1)} - {2x + 2} + {1}] = 2(2x + 3)(2x^2 + 2x + 1)

3. (x + 1)^3 - (y - 2)^3
This is the difference of two cubes. I'll let you work out the steps. I get:
(x + 1)^3 - (y - 2)^3 = (x - y + 3)(x^2 + y^2 + xy + 3x - 3y + 6)

-Dan