Solve. 64x^3-1=0

Question

Solve. 64x^3-1=0

blondisaurus · Answer

+1 not minus one >.<

anonymous · Answer

1.  $$64x^3+1=0$$2. Use sum of cubes formula:
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
3. $$(4x)^3 + (1)^3$$

$$a = 4x$$ and $$b = 1,$$

4.apply formula

$$= (4x + 1)(16x^2 - 4x + 1) $$

blondisaurus · Answer

Ohhh Okay, i get it. Thanks! :D

anonymous · Answer

4x+1)(16x^2-4x+1)=0
x=-1/4
x=$$\frac{ 1 }{ 8 }(1-i \sqrt{3})      \frac{ 1 }{ 8 }(1+i \sqrt{3})$$

blondisaurus · Answer

I dont get how you got those numbers kristen :(

blondisaurus · Answer

@kristin848 Lol come back!!

hartnn · Answer

(16x^2-4x+1)=0
for this u need quadratic formula, u know it ?

blondisaurus · Answer

Forgot it

anonymous · Answer

woops I didn't see your post lol 

so 4x+1=0
x=-1/4
 and the other two numbers you need to solve this (16x^2-4x+1) with the quadratic formula which is $$x=\frac{ -b \pm \sqrt{b^2-4ac}}{ 2a }$$

blondisaurus · Answer

How do you know which numbers are placed for c and a?

blondisaurus · Answer

Like, whats it look like when the numbers are plugged in?

anonymous · Answer

so basically, you need to use $$ax^2+bx+c$$
$$16x^2-4x+1$$
$$x=\frac{ -(-4)\pm \sqrt{16^2-4(16)(1)} }{ 2(16) }$$

blondisaurus · Answer

Thanks!