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

3 Answers
Jul 11, 2016

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

Jul 11, 2016

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!

Jul 11, 2016

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
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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")
Tony BTony B
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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