How can you find cubes of numbers without using multiplication?

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How can you find cubes of numbers without using multiplication?

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Answer 1 · 2016-10-10T20:23:32

Use logarithms and exponents.

Explanation:

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

$x^{3} = e^{3 ln x} = e^{e^{ln 3 + ln ln x}}$ $x^3 = e^(3 ln x) = e^(e^(ln 3 + ln ln x))$

Or:

$x^{3} = 10^{3 log x} = 10^{10^{log 3 + log log x}}$ $x^3 = 10^(3 log x) = 10^(10^(log 3 + log log x))$

Is that what you were looking for?

$color(white)()$
Some more explanation

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

$x^{3} = x \cdot x \cdot x$ $x^3 = x*x*x$

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 $ln$ $ln$ , which is the inverse of exponentiation base $e$ $e$ or common logarithm $log$ $log$ , which is the inverse of exponentiation base $10$ $10$ .

Note that exponentiation relates multiplication and addition like this:

$a^{m + n} = a^{m} \cdot a^{n}$ $a^(m+n) = a^m*a^n$

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

$x y = a^{{log}_{a} x + {log}_{a} y}$ $xy = a^(log_a x + log_a y)$

As an extension of this we find that:

${log}_{a} x^{n} = n {log}_{a} x$ $log_a x^n = n log_a x$

and hence:

$x^{n} = a^{n {log}_{a} x}$ $x^n = a^(n log_a x)$

We can express the multiplication $n {log}_{a} x$ $n log_a x$ in terms of exponentiation and logarithms too:

$n {log}_{a} x = a^{{log}_{a} n + {log}_{a} {log}_{a} x}$ $n log_a x = a^(log_a n + log_a log_a x)$

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

$x^{3} = e^{e^{ln 3 + ln ln x}}$ $x^3 = e^(e^(ln 3 + ln ln x))$

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

$x^{3} = 10^{10^{log 3 + log log x}}$ $x^3 = 10^(10^(log 3 + log log x))$

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

$x^{3} = e^{ln x + ln x + ln x}$ $x^3 = e^(ln x + ln x + ln x)$

$x^{3} = 10^{log x + log x + log x}$ $x^3 = 10^(log x + log x + log x)$

Answer 2 · 2016-10-11T19:17:15

Demonstration example using addition

Explanation:

Think of multiplication as addition. So explaining by example.

$2 \times 3$ $2xx3$ is 2 lots of 3 added together $\to 3 + 3 = 6$ $-> 3+3 =6$

so lets see if we can change $4^{3}$ $4^3$ into addition

We know that
$4^{3} = 4 \times 4^{2} = 4 \times (4 lots of 4) = 4 \times (4 + 4 + 4 + 4)$ $4^3=4xx4^2" "=" "4xx(4" lots of "4)=4xx(4+4+4+4)$

But $4 \times (4 + 4 + 4 + 4)$ $4xx(4+4+4+4)$ is the same as:

$14 . + 114 . + .1 4 . + 1 . 4$ $color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4$
$14 . + 114 . + .1 4 . + 1 . 4$ $color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4$
$14 . + 114 . + .1 4 . + 1 . 4$ $color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4$
$ul(color(white)(1)4color(white)(.)+color(white)(11)4color(white)(.)+color(white)(.1)4color(white)(.)+color(white)(1.)4) larr" "Add"$
$16color(white)(.)+color(white)(.)16color(white)(.)+color(white)(.)16color(white)(.)+color(white)(.)16$

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

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

Answer 3 · 2016-10-11T19:35:16

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 · 2016-10-12T06:31:50

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:

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

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

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

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

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How can you find cubes of numbers without using multiplication?

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