GoBo's Math Woes

[PuPu]

This is on p.322-323 of our textbook. It's a problem of Difference of Cubes.

First, factor out the 3 to get:

3(27x^3 - 64)

Using this property:

a^3 - b^3 = (a - b)(a^2 +ab +b^2)

We can convert it into:

3(3x - 4)[(3x)^2 - 12x + 4^2)

3(3x - 4)(9x^2 - 12x +16) = 81x^3-192

Then you set it (3x - 4) equal to 0:

3x - 4 = 0
3x = 4
x = 4/3

For the other one, (9x^2 - 12x + 16), it cannot be factored, and you have to use the Quadratic Formula.



Substitute the values of a, b, and c for:

x = [12 + √12^2 - 4(9)(16)] / 2(9)]

x = [12 - √12^2 - 4(9)(16)] / 2(9)]

Thus:

x = (12 + 12i√3)/18
x = (12 - 12i√3)/18