Question

How can you find cubes of numbers without using multiplication?

Answer 1

Use logarithms and exponents.

Explanation:

Hmmm. Well I suppose you could use addition, logarithms and exponents...

`x^3 = e^(3 ln x) = e^(e^(ln 3 + ln ln x))`

Or:

`x^3 = 10^(3 log x) = 10^(10^(log 3 + log log x))`

Is that what you were looking for?

`color(white)()`
Some more explanation

The cube of a number `x` is formed by multiplying:

`x^3 = x*x*x`

You can convert multiplication into addition using logs and exponents.

We can use any base of logarithm, but the most commonly used ones are natural logarithm `ln`, which is the inverse of exponentiation base `e` or common logarithm `log`, which is the inverse of exponentiation base `10`.

Note that exponentiation relates multiplication and addition like this:

`a^(m+n) = a^m*a^n`

So we can use it in combination with logarithm to do multiplication using addition:

`xy = a^(log_a x + log_a y)`

As an extension of this we find that:

`log_a x^n = n log_a x`

and hence:

`x^n = a^(n log_a x)`

We can express the multiplication `n log_a x` in terms of exponentiation and logarithms too:

`n log_a x = a^(log_a n + log_a log_a x)`

Hence putting `a = e` and `n = 3` we find:

`x^3 = e^(e^(ln 3 + ln ln x))`

Putting `a=10` and `n=3` we find:

`x^3 = 10^(10^(log 3 + log log x))`

Alternatively, we could use addition instead of multiplication by `3` to get the forms:

`x^3 = e^(ln x + ln x + ln x)`

`x^3 = 10^(log x + log x + log x)`

Answer 2

Demonstration example using addition

Explanation:

Think of multiplication as addition. So explaining by example.

`2xx3` is 2 lots of 3 added together `-> 3+3 =6`

so lets see if we can change `4^3` into addition

We know that
`4^3=4xx4^2" "=" "4xx(4" lots of "4)=4xx(4+4+4+4)`

But `4xx(4+4+4+4)` is the same as:

`color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4`
`color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4`
`color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4`
`ul(color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4) larr" "Add"`
`16color(white)(.)+color(white)(.)16color(white)(.)+color(white)(.)16color(white)(.)+color(white)(.)16`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`16`
`16`
`16`
`ul(16) larr" Add"`
`64`

So by using addition we have demonstrated the `4^3=64`

Answer 3

If you are happy to square numbers, add, subtract and divide by `2` then:

`x^3 = ((x^2+x)/2)^2 - ((x^2-x)/2)^2`

Explanation:

Suppose you are not happy with multiplying two arbitrary numbers but:

• You know the squares of numbers.
• You are happy to add, subtract and divide by `2`.

Note first that:

`(a+b)^2 - (a-b)^2 = (a^2+2ab+b^2) - (a^2-2ab+b^2) = 4ab`

So:

`((a+b)/2)^2 - ((a-b)/2)^2 = ab`

Put `a=x^2` and `b=x` to find:

`x^3 = ((x^2+x)/2)^2 - ((x^2-x)/2)^2`

For example:

`4^3 = ((4^2+4)/2)^2 - ((4^2-4)/2)^2`

`color(white)(4^3) = ((16+4)/2)^2 - ((16-4)/2)^2`

`color(white)(4^3) = (20/2)^2 - (12/2)^2`

`color(white)(4^3) = 10^2 - 6^2`

`color(white)(4^3) = 100 - 36`

`color(white)(4^3) = 64`

Answer 4

Here's a way to construct cubes using just addition and subtraction...

Explanation:

Here's a way to construct the sequence of cubes of positive integers without using multiplication...

Write down the first four cubes of positive integers in a line with spaces between them (leaving space on the right too)...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

In the gaps under each pair of numbers, write the difference between them to make a second line...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

`color(white)(00)7color(white)(000)19color(white)(000)37`

Add a third line consisting of the differences between each pair of numbers in the second line...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

`color(white)(00)7color(white)(000)19color(white)(000)37`

`color(white)(0000)12color(white)(000)18`

Add a fourth line containing the difference between the pair of numbers in the third line...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

`color(white)(00)7color(white)(000)19color(white)(000)37`

`color(white)(0000)12color(white)(000)18`

`color(white)(00000000)6`

Extend the fourth line by repeating the number `6` we arrived at as many times as you want more cubes (I will just do two to save typing)...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

`color(white)(00)7color(white)(000)19color(white)(000)37`

`color(white)(0000)12color(white)(000)18`

`color(white)(00000000)6color(white)(0000)color(red)(6)color(white)(0000)color(red)(6)`

Construct resulting additional numbers for the third line by adding...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

`color(white)(00)7color(white)(000)19color(white)(000)37`

`color(white)(0000)12color(white)(000)18color(white)(000)color(red)(24)color(white)(000)color(red)(30)`

`color(white)(00000000)6color(white)(0000)color(red)(6)color(white)(0000)color(red)(6)`

Repeat to get additional terms for the second line...

`1color(white)(0000)8color(white)(000)27color(white)(000)64`

`color(white)(00)7color(white)(000)19color(white)(000)37color(white)(000)color(red)(61)color(white)(000)color(red)(91)`

`color(white)(0000)12color(white)(000)18color(white)(000)color(red)(24)color(white)(000)color(red)(30)`

`color(white)(00000000)6color(white)(0000)color(red)(6)color(white)(0000)color(red)(6)`

Finally repeat for the first row to get:

`1color(white)(0000)8color(white)(000)27color(white)(000)64color(white)(00)color(red)(125)color(white)(00)color(red)(216)`

`color(white)(00)7color(white)(000)19color(white)(000)37color(white)(000)color(red)(61)color(white)(000)color(red)(91)`

`color(white)(0000)12color(white)(000)18color(white)(000)color(red)(24)color(white)(000)color(red)(30)`

`color(white)(00000000)6color(white)(0000)color(red)(6)color(white)(0000)color(red)(6)`