# How can you find cubes of numbers without using multiplication?

Oct 10, 2016

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?

$\textcolor{w h i t e}{}$
Some more explanation

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

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

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:

${a}^{m + n} = {a}^{m} \cdot {a}^{n}$

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

$x y = {a}^{{\log}_{a} x + {\log}_{a} y}$

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:

${x}^{3} = {e}^{\ln x + \ln x + \ln x}$

${x}^{3} = {10}^{\log x + \log x + \log x}$

Oct 11, 2016

#### Explanation:

Think of multiplication as addition. So explaining by example.

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

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 $4 \times \left(4 + 4 + 4 + 4\right)$ is the same as:

$\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4$
$\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4$
$\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4$
$\underline{\textcolor{w h i t e}{1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{11} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.1} 4 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{1.} 4} \leftarrow \text{ "Add}$
$16 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} 16 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} 16 \textcolor{w h i t e}{.} + \textcolor{w h i t e}{.} 16$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$16$
$16$
$16$
$\underline{16} \leftarrow \text{ Add}$
$64$

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

Oct 11, 2016

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

${x}^{3} = {\left(\frac{{x}^{2} + x}{2}\right)}^{2} - {\left(\frac{{x}^{2} - x}{2}\right)}^{2}$

#### 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:

${\left(a + b\right)}^{2} - {\left(a - b\right)}^{2} = \left({a}^{2} + 2 a b + {b}^{2}\right) - \left({a}^{2} - 2 a b + {b}^{2}\right) = 4 a b$

So:

${\left(\frac{a + b}{2}\right)}^{2} - {\left(\frac{a - b}{2}\right)}^{2} = a b$

Put $a = {x}^{2}$ and $b = x$ to find:

${x}^{3} = {\left(\frac{{x}^{2} + x}{2}\right)}^{2} - {\left(\frac{{x}^{2} - x}{2}\right)}^{2}$

For example:

${4}^{3} = {\left(\frac{{4}^{2} + 4}{2}\right)}^{2} - {\left(\frac{{4}^{2} - 4}{2}\right)}^{2}$

$\textcolor{w h i t e}{{4}^{3}} = {\left(\frac{16 + 4}{2}\right)}^{2} - {\left(\frac{16 - 4}{2}\right)}^{2}$

$\textcolor{w h i t e}{{4}^{3}} = {\left(\frac{20}{2}\right)}^{2} - {\left(\frac{12}{2}\right)}^{2}$

$\textcolor{w h i t e}{{4}^{3}} = {10}^{2} - {6}^{2}$

$\textcolor{w h i t e}{{4}^{3}} = 100 - 36$

$\textcolor{w h i t e}{{4}^{3}} = 64$

Oct 12, 2016

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)...

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

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

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$

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

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$

$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18$

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

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$

$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18$

$\textcolor{w h i t e}{00000000} 6$

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)...

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$

$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18$

$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37$

$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18 \textcolor{w h i t e}{000} \textcolor{red}{24} \textcolor{w h i t e}{000} \textcolor{red}{30}$

$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$

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

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37 \textcolor{w h i t e}{000} \textcolor{red}{61} \textcolor{w h i t e}{000} \textcolor{red}{91}$

$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18 \textcolor{w h i t e}{000} \textcolor{red}{24} \textcolor{w h i t e}{000} \textcolor{red}{30}$

$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$

Finally repeat for the first row to get:

$1 \textcolor{w h i t e}{0000} 8 \textcolor{w h i t e}{000} 27 \textcolor{w h i t e}{000} 64 \textcolor{w h i t e}{00} \textcolor{red}{125} \textcolor{w h i t e}{00} \textcolor{red}{216}$

$\textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{000} 19 \textcolor{w h i t e}{000} 37 \textcolor{w h i t e}{000} \textcolor{red}{61} \textcolor{w h i t e}{000} \textcolor{red}{91}$

$\textcolor{w h i t e}{0000} 12 \textcolor{w h i t e}{000} 18 \textcolor{w h i t e}{000} \textcolor{red}{24} \textcolor{w h i t e}{000} \textcolor{red}{30}$

$\textcolor{w h i t e}{00000000} 6 \textcolor{w h i t e}{0000} \textcolor{red}{6} \textcolor{w h i t e}{0000} \textcolor{red}{6}$