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

Answer:

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

Answer:

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

Answer:

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")#
Tony 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#