@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!
–
Lu Chee Ket
·
1 year, 8 months ago

Log in to reply

@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.
–
Lu Chee Ket
·
1 year, 8 months ago

Log in to reply

(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.
–
Lu Chee Ket
·
1 year, 9 months ago

Log in to reply

(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!
–
Lu Chee Ket
·
1 year, 9 months ago

## Comments

Sort by:

TopNewestexplain me the answer – Sandeep Sunny · 1 year, 1 month ago

Log in to reply

I'm not sure this is solvable. Wolfram Alpha doesn't give a

nicesolution.On the other hand, \( x^3 + y^3 + 30xy = 1000 \) is solvable. Did you mean this? – Siddhartha Srivastava · 1 year, 9 months ago

Log in to reply

Please explain your opinion about x^3 + y^3 + 30 x y = 2000, for x + y. – Lu Chee Ket · 1 year, 9 months ago

Log in to reply

Wolfram Alpha to check if it has a solution.Wolfram Alpha shows that it does, but the solutions are complex. – Siddhartha Srivastava · 1 year, 9 months ago

I usedLog in to reply

– Lu Chee Ket · 1 year, 8 months ago

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!Log in to reply

– Lu Chee Ket · 1 year, 8 months ago

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.Log in to reply

(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. – Lu Chee Ket · 1 year, 9 months ago

Log in to reply

(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! – Lu Chee Ket · 1 year, 9 months ago

Log in to reply