Question

Three numbers are in the ratio 2:3:4. The sum of their cubes is 0.334125. How do you find the numbers?

Answer 1

The 3 numbers are: `0.3, 0.45, 0.6`

Explanation:

The question says there are three numbers but with a specific ratio. What that means is that once we pick one of the numbers, the other two are known to us through the ratios. We can therefore replace all 3 of the numbers with a single variable:

`2:3:4 implies 2x:3x:4x`

Now, no matter what we choose for `x` we get the three numbers in the ratios specified. We are also told the sum of the cubes of these three numbers which we can write:

`(2x)^3+(3x)^3+(4x)^3 = 0.334125`

distributing the powers across the factors using `(a*b)^c = a^c b^c` we get:

`8x^3+27x^3+64x^3 = 99x^3 = 0.334125`

`x^3 = 0.334125/99 = 0.003375`

`x = root(3)0.003375 = 0.15`

So the 3 numbers are:

`2*0.15, 3*0.15, 4*0.15 implies 0.3, 0.45, 0.6`

Answer 2

The nos. are, `0.3, 0.45, and, 0.6`.

Explanation:

Reqd. nos. maintain ratio `2:3:4`. Therefore, let us take the reqd. nos. to be `2x, 3x, and, 4x.`

By what is given, `(2x)^3+(3x)^3+(4x)^3=0.334125`

`rArr 8x^3+27x^3+64x^3=0.334125`

`rArr 99x^3=0.334125`

`rArr x^3=0.334125/99=0.003375=(0.15)^3...................(1)`

`rArr x=0.15`

So, the nos. are, `2x=0.3, 3x=0.45, and, 4x=0.6`.

This soln. is in `RR`, but, for that in `CC`, we can solve eqn.(1) as under :-

`x^3-0.15^3=0 rArr (x-0.15)(x^2+0.15x+0.15^2)=0`

`rArr x=0.15, or, x={-0.15+-sqrt(0.15^2-4xx1xx0.15^2)}/2`

`rArr x=0.15, x={-0.15+-sqrt(0.15^2xx-3)}/2`

`rArr x=0.15, x=(-0.15+-0.15*sqrt3*i)/2`

`rArr x=0.15, x=(0.15){(-1+-sqrt3i)/2}`

`rArr x=0.15, x=0.15omega, x=0.15omega^2`

I leave it to you to verify if complex roots satisfy the given cond. - hoping that you'll enjoy it!

Answer 3

Slightly different approach.

`"First number: "2/9a->2/9xx27/20 = 3/10 -> 0.3`

`"Second number: "3/9a-> 3/9xx27/20=9/20->0.45`

`"Third number: "4/9a->4/9xx27/20=3/5->0.6`

Explanation:

We have a ratio which is splitting the whole of something into proportions.

Total number of parts `=2+3+4 = 9" parts"`
Let the whole thing be `a` ( for all )

Then `a=2/9a+3/9a+4/9a`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We are told that the sum of their cubes is `0.334125`

Note that `0.334125 = 334125/1000000 -= 2673/8000`

( aren't calculators are wonderful!)

So `(2/9a)^3+(3/9a)^3+(4/9a)^3=2673/8000`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`8/729a^3+27/729a^3 +64/729a^3=2673/8000`

Factor out the `a^3`

`a^3(8/729+27/729 +64/729)=2673/8000`

`a^3=2673/8000xx729/99`

`a^3=19683/8000`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Looking for cubed numbers")`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`a^3 =(3^3xx3^3xx3^3 )/(10^3xx2^3)`

Take the cube root of both sides

`a=(3xx3xx3)/(10xx2) = 27/20`

`color(white)(2/2)`

`color(brown)("So the numbers are:")`

`"First number: "2/9a->2/9xx27/20 = 3/10 -> 0.3`

`"Second number: "3/9a-> 3/9xx27/20=9/20->0.45`

`"Third number: "4/9a->4/9xx27/20=3/5->0.6`