This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
(x + y)^3 - 3 x y (x + y) + 30 x y - 2000 = 0
If z = x + y then z^3 - 3 x y z + 30 x y - 2000 = 0
Do you think solvable? z^3 + p z + q = 0 is solvable because p and q are constant not related to z as an equation rather than an identity.
Here, p = - 3 x y and q = 30 x y - 2000. Therefore, we come back to two unknowns require two distinct equations to be solvable for fixed values. We can only write q = - 10 p - 2000 and cannot tell what p and q can do for x and y. Not like derivation of cubic formula using the feature of identity (always equal), unless 30 x y - 2000 is changed into - (x^3 + y^3) to form the identity, it could not be having the same fate that x and y are intersected for fixed values. If 30 x y - 2000 can be a constant, then x + y is solvable. The whole thing x^3 + y^3 + 30 x y= 2000 does not help at solving a cubic equation as values for coefficient p and q cannot be known.
If not mistaken , then I think the question is not solvable and hence x + y cannot be determined unless there is descriptions about integer values or something related. See x^3 + y^3 + 30 x y = 2000 again, some x can have some y but not fixed, not likely to be constant sum unless it does . Hence x + y cannot be fixed and therefore not solvable!
@Siddhartha Srivastava
–
I plugged in a value to y to solve for x, when I changed for another value, (x + y) varied. If not mistaken, then such a check should have convinced to me that x + y is not constant with arbitrary thousands. Therefore, x + y could not be solved generally. Thanks very much for answering to me. You can check whether I am careless at making incorrect calculations. I have just checked that x+ y cannot be constant for 2000.
@Siddhartha Srivastava
–
True. Wolfram Alpha gives the same conclusion while being more complete. Giving answer in term of x is also considered as solving is a refreshed idea. Trying the best for what can be reached. Thanks very much for introducing this!
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
(x + y)^3 - 3 x y (x + y) + 30 x y - 2000 = 0
If z = x + y then z^3 - 3 x y z + 30 x y - 2000 = 0
Do you think solvable? z^3 + p z + q = 0 is solvable because p and q are constant not related to z as an equation rather than an identity.
Here, p = - 3 x y and q = 30 x y - 2000. Therefore, we come back to two unknowns require two distinct equations to be solvable for fixed values. We can only write q = - 10 p - 2000 and cannot tell what p and q can do for x and y. Not like derivation of cubic formula using the feature of identity (always equal), unless 30 x y - 2000 is changed into - (x^3 + y^3) to form the identity, it could not be having the same fate that x and y are intersected for fixed values. If 30 x y - 2000 can be a constant, then x + y is solvable. The whole thing x^3 + y^3 + 30 x y= 2000 does not help at solving a cubic equation as values for coefficient p and q cannot be known.
If not mistaken , then I think the question is not solvable and hence x + y cannot be determined unless there is descriptions about integer values or something related. See x^3 + y^3 + 30 x y = 2000 again, some x can have some y but not fixed, not likely to be constant sum unless it does . Hence x + y cannot be fixed and therefore not solvable!
(x + y)^3 - 3 x y (x + y) - (x^3 + y^3) = 0 {An identity taken for solving cubic equation.}
With p = -3 x y and q = - (x^3 + y^3)
y = p/ (-3 x) substitutes to q = - (x^3 + y^3) yields
-q = x^3 - p^3/ (27 x^3)
x^6 + q x^3 - p^3/ 27 = 0
x^2 = -q/ 2 +/- Sqrt [q^2/ 4 + p^3/ 27]
x = +/- Sqrt {-q/ 2 +/- Sqrt [q^2/ 4 + p^3/ 27]}
Since x and y here are symmetrical, one of the solution to cubic equation with real coefficients is having alpha as:
Taking x = Sqrt {-q/ 2 + Sqrt [q^2/ 4 + p^3/ 27]} and y = Sqrt {-q/ 2 - Sqrt [q^2/ 4 + p^3/ 27]}
Hence, for z^3 + p z + q = 0,
z = x + y = Sqrt {-q/ 2 + Sqrt [q^2/ 4 + p^3/ 27]} + Sqrt {-q/ 2 - Sqrt [q^2/ 4 + p^3/ 27]}
This is the derivation of the cubic formula.
I'm not sure this is solvable. Wolfram Alpha doesn't give a nice solution.
On the other hand, x3+y3+30xy=1000 is solvable. Did you mean this?
Log in to reply
All right, x + y = 10 for x^3 + y^3 + 30 x y = 1000 as you meant. This was what I meant not likely, but a coincidence.
Please explain your opinion about x^3 + y^3 + 30 x y = 2000, for x + y.
Log in to reply
I used Wolfram Alpha to check if it has a solution.Wolfram Alpha shows that it does, but the solutions are complex.
Log in to reply
I plugged in a value to y to solve for x, when I changed for another value, (x + y) varied. If not mistaken, then such a check should have convinced to me that x + y is not constant with arbitrary thousands. Therefore, x + y could not be solved generally. Thanks very much for answering to me. You can check whether I am careless at making incorrect calculations. I have just checked that x+ y cannot be constant for 2000.
True. Wolfram Alpha gives the same conclusion while being more complete. Giving answer in term of x is also considered as solving is a refreshed idea. Trying the best for what can be reached. Thanks very much for introducing this!
explain me the answer