How can you find cubes of numbers without using multiplication?

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)#

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#

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#

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)#