how to apply sum of two cubes here

A

allegansveritatem

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Here is a system of equations that is to be solved using substitution:
cube sum.PNG

Here is what I tried to do with it:
cube sum 2.PNG
Somehow the fact that there is a cooeficient that is not a cube makes me hesitate to try anything with the difference of two cubes formula. Is said formula even applicable here. If so how, if not, what would be the way forward. Seems to me I am missing the primrose path here and have wandered into the bramble patch.
 
Substitution? That's crazy. Elimination is the way to go.
Having said that you clearly stated that you have to use substitution and you did try.
The first problem I saw with your work is that you have a mistake on your first line (sorry). Please fix that and see where you can get from there.
 
OK, your 2nd attempt is better.
To answer you question (a^3 + b^3) = (a+b)(a^2 -ab +b^2). Your a is cuberoot(21) and your b is 1.
My question is why do you want to do that? Why not just say that 21x^3 = -1 so x^3= -1/21 and solve for x by computing the cubert of both sides?
 
Here is my approach:
\(\begin{gathered}
\left\{ \begin{gathered}
6{x^3} - {y^3} = 1 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
24{x^3} - 4{y^3} = 4 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right.\\ \to 27{x^3} = 9 \hfill \\
3x = \sqrt[3]{9} \to x = \frac{{\sqrt[3]{9}}}{3} \hfill \\
\end{gathered} \)
Can you find \(y~?\)
 
pka said:
Here is my approach:
\(\begin{gathered}
\left\{ \begin{gathered}
6{x^3} - {y^3} = 1 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
24{x^3} - 4{y^3} = 4 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right.\\ \to 27{x^3} = 9 \hfill \\
3x = \sqrt[3]{9} \to x = \frac{{\sqrt[3]{9}}}{3} \hfill \\
\end{gathered} \)
Can you find \(y~?\)
Click to expand...
The OP requested to solve this problem by substitution.
 
You know that 6x^3 - y^3 = 1. So 6x^3 = 1 + y^3.
The 2nd ens is 3x^3 + 4y^3 = 5 or 6x^3 + 8 y^3 = 10.
Now substitute.
 
Jomo said:
The OP requested to solve this problem by substitution.
Click to expand...
So what what? Why not show a more practical way of solving the problem.
BTW the problem is clearly designed to be solved by elimination.
 
pka said:
So what what? Why not show a more practical way of solving the problem.
BTW the problem is clearly designed to be solved by elimination.
Click to expand...
Yes, and if you read my initial post I said that this is clearly an elimination problem
 
Because elimination is not generally applicable to non-linear equations, I see no objection to saying to solve this equation by substitution. In any case, that seems to be what is required. Following the OP's method but avoiding the sign error

[MATH]6x^3 - y^3 = 1 \implies y^3 = 6x^3 - 1.[/MATH]
You do not need to find y because you are going to substitute for y^3 rather than y.

[MATH]3x^3 + 4y^3 = 5 \implies 5 = 3x^3 + 4(6x^3 - 1) = 27x^3 - 4 \implies \\ 27x^3 = 9 \implies x^3 = \dfrac{1}{3} \implies y^3 = 6 * \dfrac{1}{3} - 1 = 1.[/MATH]Let's check.

[MATH]3 * \dfrac{1}{3} + 4 * 1 = 1 + 4 = 5 \ \checkmark.[/MATH]
[MATH]6 * \dfrac{1}{3} - 1 = 2 - 1 = 1 \ \checkmark.[/MATH]
[MATH]x^3 = \dfrac{1}{3} \implies x = \sqrt[3]{\dfrac{1}{3}} \text { and } y^3 = 1 \implies y = 1.[/MATH]
EDIT: This would also be a nice problem to introduce u-substitutions.
 
Last edited:
Jomo said:
OK, your 2nd attempt is better.
To answer you question (a^3 + b^3) = (a+b)(a^2 -ab +b^2). Your a is cuberoot(21) and your b is 1.
My question is why do you want to do that? Why not just say that 21x^3 = -1 so x^3= -1/21 and solve for x by computing the cubert of both sides?
Click to expand...
Yes, the instruction is to use substitution. I will have to study this post and the others later today and get back to you. Thanks
 
pka said:
Here is my approach:
\(\begin{gathered}
\left\{ \begin{gathered}
6{x^3} - {y^3} = 1 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
24{x^3} - 4{y^3} = 4 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right.\\ \to 27{x^3} = 9 \hfill \\
3x = \sqrt[3]{9} \to x = \frac{{\sqrt[3]{9}}}{3} \hfill \\
\end{gathered} \)
Can you find \(y~?\)
Click to expand...
Thanks for reply, I will get back to you
 
JeffM said:
Because elimination is not generally applicable to non-linear equations, I see no objection to saying to solve this equation by substitution. In any case, that seems to be what is required. Following the OP's method but avoiding the sign error

[MATH]6x^3 - y^3 = 1 \implies y^3 = 6x^3 - 1.[/MATH]
You do not need to find y because you are going to substitute for y^3 rather than y.

[MATH]3x^3 + 4y^3 = 5 \implies 5 = 3x^3 + 4(6x^3 - 1) = 27x^3 - 4 \implies \\ 27x^3 = 9 \implies x^3 = \dfrac{1}{3} \implies y^3 = 6 * \dfrac{1}{3} - 1 = 1.[/MATH]Let's check.

[MATH]3 * \dfrac{1}{3} + 4 * 1 = 1 + 4 = 5 \ \checkmark.[/MATH]
[MATH]x^3 = \dfrac{1}{3} \implies x = \sqrt[3]{\dfrac{1}{3}} \text { and } y^3 = 1 \implies y = 1.[/MATH]
EDIT: This would also be a nice problem to introduce u-substitutions.
Click to expand...
I will study this later today and get back to you. Thanks
 
Does the ORIGINAL PROBLEM say anything about difference or sum of cubes? Because that would indeed by crazy. Or did you look up difference and sum of cubes and find that the formula applies even when the coefficients are not perfect cubes. That is true, but the formulas are not relevant to solving this problem as jomo, pka, and I have all shown.
 
Just at a guess I would say that the substitution part merely refers to something like:

Define [math]m = x^3[/math] and [math]n = y^3[/math]. Then solve
[math]\cases{6m - n = 1\\ 3m + 4n = 5}[/math]
-Dan
 
topsquark said:
Just at a guess I would say that the substitution part merely refers to something like:

Define [math]m = x^3[/math] and [math]n = y^3[/math]. Then solve
[math]\cases{6m - n = 1\\ 3m + 4n = 5}[/math]
-Dan
Click to expand...
I get it. I should have been able to figure that out. Thanks
pka said:
Here is my approach:
\(\begin{gathered}
\left\{ \begin{gathered}
6{x^3} - {y^3} = 1 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
24{x^3} - 4{y^3} = 4 \hfill \\
3{x^3} + 4{y^3} = 5 \hfill \\
\end{gathered} \right.\\ \to 27{x^3} = 9 \hfill \\
3x = \sqrt[3]{9} \to x = \frac{{\sqrt[3]{9}}}{3} \hfill \\
\end{gathered} \)
Can you find \(y~?\)
Click to expand...
I worked this out two ways--one of the ways suggested above (with a twist in that I start by finding Y first) and then I go back and redo what I started with.
 
I worked this out two ways--one of the ways suggested above (with a twist in that I start by finding Y first) and then I go back and redo what I started with:
1)cube5.PNG

and 2)cube3.PNG

Everything seems to agree with everything else so....
 
Your work is fine.
I just think that after finding what x^3 equals then founding what x equals just to cube it and arrive back to x^3 is inefficient. You need to see the future. If you have x^3 in the 2nd equation, then if you find what x^3 equals in the 1st equation then you are done--that is no need to find x.
 
Jomo said:
Your work is fine.
I just think that after finding what x^3 equals then founding what x equals just to cube it and arrive back to x^3 is inefficient. You need to see the future. If you have x^3 in the 2nd equation, then if you find what x^3 equals in the 1st equation then you are done--that is no need to find x.
Click to expand...
I will have to get back into what we were doing with this problem before I can get what you are saying here. I will check into this later. I have a feeling where you are going here but I will have to go back to what is being referenced.
 
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