MHB What Are Some Tricks To Calculate n^3?

  • Thread starter Thread starter susanto3311
  • Start date Start date
AI Thread Summary
To calculate n^3, one effective method is to multiply the number by itself twice, such as 17^3 = 17 * 17 * 17. A practical approach involves rounding the number to simplify calculations, like using 20 for 17, to make the multiplication easier. The binomial expansion formula can also be applied, for instance, (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, which simplifies calculations for numbers like 31. Additionally, the difference of cubes formula can be utilized to break down the calculation further. Overall, these methods enhance the ease of calculating cubic values without a calculator.
susanto3311
Messages
73
Reaction score
0
hi all..

how to easy calculate this problem :

17^3 = ...
13^3 =...
31^3= ...

do you have simple method to figure it out?
 
Mathematics news on Phys.org
Well, you just have to multiply 17 with itself twice, that is, 17^3 = 17 * 17 * 17... then you can use whatever method you want to do it "easily"/quickly. For instance if I did not have access to a calculator what I would do is start with one multiplication like this:

17 * 17 = (20 - 3) * (20 - 3) = 20 * 20 - 20 * 3 - 20 * 3 + 3 * 3 = 400 - 60 - 60 + 9 = 289

Then do the second multiplication as:

17 * 17 * 17 = 289 * 17 = (300 - 11) * (20 - 3) = 300 * 20 - 300 * 3 - 11 * 20 + 11 * 3 = 6000 - 900 - 220 + 33 = 4913

And that's it. Same goes for 13 or 31 ( the trick here is working out what you should round each number to make sure you're actually making the problem simpler and not harder, multiples of 10 or 100 are usually a good bet). If you have a calculator you can just punch it in though. But really it's just multiplication.
 
You can also use the binomial expansion:

$$(a+b)^3=a^3+3a^2b+4ab^2+b^3$$

So, for example, you would find:

$$31^3=(30+1)^3=30^3+3\cdot30^2\cdot1+3\cdot30\cdot1^2+1^3=27000+2700+90+1=29791$$

Or you could use a difference of cubes formula:

$$a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$$

So, for example, we could write:

$$31^3=\left(31^2-30^3\right)+30^3=(31-30)\left(31^2+31\cdot30+30^2\right)+30^3=961+930+900+27000=29791$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top