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

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:

$2 : 3 : 4 \implies 2 x : 3 x : 4 x$

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:

${\left(2 x\right)}^{3} + {\left(3 x\right)}^{3} + {\left(4 x\right)}^{3} = 0.334125$

distributing the powers across the factors using ${\left(a \cdot b\right)}^{c} = {a}^{c} {b}^{c}$ we get:

$8 {x}^{3} + 27 {x}^{3} + 64 {x}^{3} = 99 {x}^{3} = 0.334125$

${x}^{3} = \frac{0.334125}{99} = 0.003375$

$x = \sqrt[3]{0.003375} = 0.15$

So the 3 numbers are:

$2 \cdot 0.15 , 3 \cdot 0.15 , 4 \cdot 0.15 \implies 0.3 , 0.45 , 0.6$

Jul 11, 2016

The nos. are, $0.3 , 0.45 , \mathmr{and} , 0.6$.

Explanation:

Reqd. nos. maintain ratio $2 : 3 : 4$. Therefore, let us take the reqd. nos. to be $2 x , 3 x , \mathmr{and} , 4 x .$

By what is given, ${\left(2 x\right)}^{3} + {\left(3 x\right)}^{3} + {\left(4 x\right)}^{3} = 0.334125$

$\Rightarrow 8 {x}^{3} + 27 {x}^{3} + 64 {x}^{3} = 0.334125$

$\Rightarrow 99 {x}^{3} = 0.334125$

$\Rightarrow {x}^{3} = \frac{0.334125}{99} = 0.003375 = {\left(0.15\right)}^{3.} \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$

$\Rightarrow x = 0.15$

So, the nos. are, $2 x = 0.3 , 3 x = 0.45 , \mathmr{and} , 4 x = 0.6$.

This soln. is in $\mathbb{R}$, but, for that in $\mathbb{C}$, we can solve eqn.(1) as under :-

${x}^{3} - {0.15}^{3} = 0 \Rightarrow \left(x - 0.15\right) \left({x}^{2} + 0.15 x + {0.15}^{2}\right) = 0$

$\Rightarrow x = 0.15 , \mathmr{and} , x = \frac{- 0.15 \pm \sqrt{{0.15}^{2} - 4 \times 1 \times {0.15}^{2}}}{2}$

$\Rightarrow x = 0.15 , x = \frac{- 0.15 \pm \sqrt{{0.15}^{2} \times - 3}}{2}$

$\Rightarrow x = 0.15 , x = \frac{- 0.15 \pm 0.15 \cdot \sqrt{3} \cdot i}{2}$

$\Rightarrow x = 0.15 , x = \left(0.15\right) \left\{\frac{- 1 \pm \sqrt{3} i}{2}\right\}$

$\Rightarrow x = 0.15 , x = 0.15 \omega , x = 0.15 {\omega}^{2}$

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.

$\text{First number: } \frac{2}{9} a \to \frac{2}{9} \times \frac{27}{20} = \frac{3}{10} \to 0.3$

$\text{Second number: } \frac{3}{9} a \to \frac{3}{9} \times \frac{27}{20} = \frac{9}{20} \to 0.45$

$\text{Third number: } \frac{4}{9} a \to \frac{4}{9} \times \frac{27}{20} = \frac{3}{5} \to 0.6$

Explanation:

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

Total number of parts $= 2 + 3 + 4 = 9 \text{ parts}$
Let the whole thing be $a$ ( for all )

Then $a = \frac{2}{9} a + \frac{3}{9} a + \frac{4}{9} a$
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We are told that the sum of their cubes is $0.334125$

Note that $0.334125 = \frac{334125}{1000000} \equiv \frac{2673}{8000}$

( aren't calculators are wonderful!)

So ${\left(\frac{2}{9} a\right)}^{3} + {\left(\frac{3}{9} a\right)}^{3} + {\left(\frac{4}{9} a\right)}^{3} = \frac{2673}{8000}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\frac{8}{729} {a}^{3} + \frac{27}{729} {a}^{3} + \frac{64}{729} {a}^{3} = \frac{2673}{8000}$

Factor out the ${a}^{3}$

${a}^{3} \left(\frac{8}{729} + \frac{27}{729} + \frac{64}{729}\right) = \frac{2673}{8000}$

${a}^{3} = \frac{2673}{8000} \times \frac{729}{99}$

${a}^{3} = \frac{19683}{8000}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Looking for cubed numbers}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${a}^{3} = \frac{{3}^{3} \times {3}^{3} \times {3}^{3}}{{10}^{3} \times {2}^{3}}$

Take the cube root of both sides

$a = \frac{3 \times 3 \times 3}{10 \times 2} = \frac{27}{20}$

$\textcolor{w h i t e}{\frac{2}{2}}$

$\textcolor{b r o w n}{\text{So the numbers are:}}$

$\text{First number: } \frac{2}{9} a \to \frac{2}{9} \times \frac{27}{20} = \frac{3}{10} \to 0.3$

$\text{Second number: } \frac{3}{9} a \to \frac{3}{9} \times \frac{27}{20} = \frac{9}{20} \to 0.45$

$\text{Third number: } \frac{4}{9} a \to \frac{4}{9} \times \frac{27}{20} = \frac{3}{5} \to 0.6$