Factor Completely: 8x^3-27=0

Question

anonymous · Answer

This problem is factorable by recognizing it as a difference of cubes. The formula for factoring a difference of cubes is:
$$a^{3} - b^{3} = (a-b)(a^{2}+ab+b^{2})$$
So all you need to do is find what your "a" and "b" are and plug them into this formula. Think you can look at your problem and see what a and b are?

anonymous · Answer

8 is A and -27 is B?

anonymous · Answer

Well, the first term you have is 8x^3, which would be your a^3 in this case. So to get a, you need to find the cube root of 8x^3. The same thing for b. The -27 is actually b^3. So you would need to take the cube root of that to find b. Make sense?

anonymous · Answer

Yeah. So the cube root of 8^3 would be 8 and -27 would be 3 right?

anonymous · Answer

or sorry -3

anonymous · Answer

Yes, -3, exactly. Now for 8x^3, I need to find a number that when multiplied 3 times gives us 8. The x^3 also has to be considered. If I have x^3, then the cube root of that is simply x, since x*x*x = x^3. So what would the cube root of 8 be?

anonymous · Answer

2

anonymous · Answer

Yep. So that makes the cube root of 8x^3 equal to 2x. So if a = 2x and b = -3, we can plug them into the formula:
$$(2x-3)((2x)^{2}+(2x)(3)+(3)^{2})$$ So now you would just simplify that. Gotta head out now. Good luck :)