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What is NxNxN?

Updated: 9/24/2023
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Q: What is NxNxN?
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What is the number sequence 1 8 27 64?

Cubes of digits 1 - 4n3. (nxnxn) 13 = 1, 23 = 8, 33 = 27...1x1x1 = 12x2x2 = 83x3x3 = 274x4x4 = 645x5x5 = 1256x6x6 = 216


What is cube of number?

if you are talking about squaring then that means to multiplt the base number by iself like this 72= 7*7= 49 But we are not talking about squaring. The cube of a number N, is NxNxN so 5 cubed is 5x5x5=125


If you have the volume of a cube how do you calculate the dimensions?

Take the cube root of that number. For example, volume is 8 square inches and the cube root of 8 is 2. In case you forgot, if N is the Cube root of a number M then NxNXN=M


Is a negative timesa negative times a negative a positive or a negative?

A negative times a negative times a negative (NxNxN) is a : Negative!!! Negative times a negative is positive. A positive times a negative is a negative. Remember if you are multiplying an odd number of negatives irrespective of the number of positives, it is always a negative. If you are multiplying an even number of negatives then the answer is a positive. EX: -2 x -1 x -2 ( 3 negatives so answer is negative) 2 x -2 -4 -3x-5x-1x-1x2x1 ( 4 negatives so answer is positive) 15x1x2 30


Why does any number to the zero power equal one?

Here is why any number to the zero power equals one. Consider this. a^b. it is natural to restrict a > 0, but we'll only assume that number b is any real number. We'll use the natural exponential function defined by the derivative of the exponential function. Now we have a^r=e^rln(a). And we know that e^rln(a)=e^((ln(a))^r), where a >0 and r is in the domain of all real numbers negative infinity to infinity. We can apply this definition to any number a to any power r. Particularly, a^0. By the provided definition, a^0=e^(0*ln(a))=e^0=1. Furthermore, a^1=e^(1*ln(a))=e(ln(a))=a. And a^2=(e^(ln(a))^2)=a^2. ---------------------------------- Here is a simpler approach: In general, a^n/a^m = a^(n-m) and a/a = 1. We can use these facts to prove that x^0 = 1 so long as x isn't 0. First, state the obvious: 1 = 1 Next, since any non-zero number divided by itself is one: 1 = a^n/a^n (It doesn't change how the equation looks, but for the sake of being thorough, you could subsitute (a^n/a^n) in place of 1 in the original equation.) Then, since dividing like bases requires that you subtract their exponents: a^n/a^n = a^(n-n) = a^0 Substitute (a^0) in for (a^n/a^n) and you obtain: a^0 = 1 There are two reasons "a" cannot be 0 in this proof: firstly, raising 0 to non-zero powers would still result in zero, so "a" being 0 would cause division by zero in the initial theorems we used, and secondly, 0^0 is considered undefined in itself.