1. ## Confused

number 8 is written as 3 components sum like that, where the first 2 cube amount(a^3+b^3.that means this sum is the lowest), plus 3rd component (C), multiplied by 9, is the lowest.

It is also known, that the first component is 2 times greater than second one. Find those components.

$8 = a + b + c$

$(a^3 + b^3 + c) * 9 > 0$

$2a = b$

I do get $81a^3 - 27a +64 = 0$ and I believe this is nonse, what am I doing wrong? ;\

The answer is $2; 1; 5.$

Thank you.

2. It's should be: a=2b (It is also known, that the first component is 2 times greater than second one)

I don't understand this line:

"Number 8 is written as 3 component's sum, when the first two compenents third degree sum, plus 3rd component, multiplied by 9,### is the lowest###."

3. Originally Posted by Ellla
Number 8 is written as 3 component's sum, where the first two compenents third degree sum, plus 3rd component, multiplied by 9, is the lowest.

It is also known, that the first component is 2 times greater than second one. Find those components.

$8 = a + b + c$

$(a^3 + b^3 + c) * 9 > 0$

$2a = b$

I do get $81a^3 - 27a +64 = 0$ and I believe this is nonse, what am I doing wrong? ;\

The answer is $2; 1; 5.$

Thank you.
If a, b, c are positive integers

$a+b+c=8$

$\left(a^3+b^3+c\right)9=minimum.$

$a=2b\Rightarrow\ 3b+c=8\Rightarrow\ c=8-3b$

$\left[(2b)^3+b^3+8-3b\right]9=min$

$\left(8b^3+b^3+8-3b\right]9=min$

$81b^3-27b+72=min$

$b=1$ will minimize this, from which you can find a, c.

4. Well, number 8 is written as 3 components sum like that, where the first 2 cube amount(a^3+b^3.that means this sum is the lowest), plus 3rd component (C), multiplied by 9, is the lowest.

5. Sorry, I might have confused you ;\

Archie Meade, Im very thankful for your work, but would you mind telling me how do I solve that $81a^3 - 3b + 8 = 0$ ?

6. Originally Posted by Ellla
Sorry, I might have confused you ;\

Archie Meade, Im very thankful for your work, but would you mind telling me how do I solve that $81b^3 - 3b + 8 = 0$ ?
I felt that "lowest" may have meant "smallest possible" from the available integers.

If "b" is not an integer, then you have to use a root-finding method for cubic polynomials.

If "b" is a positive integer, and we are looking for a minimum, then the $81b^3$ term dominates,
hence by inspection "b" is 1,
since higher values of "b" will give far greater results.
I think that's the context of the question, otherwise "b" would be a tiny fraction.

7. Alright, thanks.