Any easy-to-understand formulas for approximating cube roots?

Question

I can figure them pretty closely sometimes by cubing certain easy-to-multiply numbers and tinkering with them until the product is pretty close to the number I'm trying to find the cube root to. I know the values for 10 cubed, 15 cubed, 20 cubed, 25 cubed, 30 cubed, and and a few others, or can figure them pretty fast-- and that helps if the number I'm trying to find the cube root for is close to one of those. But getting a close answer that way is time-consuming, and is never quite right. The Internet shows two different formulas, but doesn't explain on the level of "an idiot's guide to cube roots", which is what I need! No particular reason. With minimal math background, I've become fascinated with working math problems lately.

Answer 1

Here is an easy way to approximate a cube root if you already have a cube that is fairly close to your answer. Lets assume that you want to find the cube root of A. Lets further assume that A is "almost" a cube, that is, there is a B such that A = B^3 + C, where C is much smaller than A.

If this is true, then the cube root of A is approximately = B + C/(3B^2)

Here is an example. Lets say we want to find the cube root of 65. We know that 4 cubed is 64, so we have the following:

65 = 4^3 + 1

Then the cube root of 65 will be approximately 4 + 1/(3(4^2))
or 4 + 1/48 which is 4.0208

The actual cube root of 65 is approximately 4.0207

Answer 2

Not really. The square root routine is really a take off on (a+b)^2 = appx a^2 +2 ab if a>>b.
I suppose you can do the same for
(a+b)^3 = a^3+3a^2b when a>>>b.
In any account, you need to know some cube roots, and your efforts are most laudable.

Answer 3

ok it truly is assuming everythings under the sq. root sign for the 1st 3 for the final 2 assume 3? is cubed root and not thrice the sq. root c) ?a^2 is the comparable as |a| comparable applies for d) e) separate ?40 9*?f^4 then remedy 7*f^2 f) 3 (3*3*3=27) g)10 h) x (x*x*x=x^3) i) separate lower back so 3?8*3?d^3 is two*d or 2d j) 3?sixty 4*3?f^6*3?g^3 is 4*f^2*g 4(f^2)g