it's a math word problem.. please help me.. i've been trying to answer it for 5 days now.. T-T?

Question

please help! I'm an LOA student this semester because of an operation. but i'm still trying to study so my brain would be ready when i come back to school..

WORK PROBLEM:
Find four numbers such that the sum of the first, third, and fourth exceeds the second by 8; the sum of the squares of the first and the second exceeds the sum of the squares of the third and fourth by 36; the sum of the products of the first and second, third and fourth is 42 (meaning: [(1st x 2nd) + (3rd x 4th) = 42]); the cube of the first equals the sum of the cubes of the second, third, and fourth.

thanks in advance!

Answer 1

Let the numbers be A, B, C, D.

Given:
(1) A + C + D = B + 8
(2) A^2 + B^2 = C^2 + D^2 + 36
(3) AB + CD = 42
(4) A^3 = B^3 + C^3 + D^3

From (1) :
A - B = 8 - (C + D)
Squaring gives :
(5) (A - B)^2 = 64 - 16(C + D) + (C + D)^2

From (2),
(A - B)^2 + 2AB = (C + D)^2 - 2CD + 36
(A - B)^2 = (C + D)^2 - 2(AB + CD) + 36
Substituting (3) into this gives :
(6) (A - B)^2 = (C + D)^2 - 48

Equating (5) with (6) gives :
64 - 16(C + D) + (C + D)^2 = (C + D)^2 - 48
Therefore, C + D = 7

Substituting this into (1) gives :
A + 7 = B + 8, or, A - B = 1

From (4),
A^3 - B^3 = C^3 + D^3
(A - B)(A^2 + AB + B^2) = (C + D)(C^2 - CD + D^2)
(A - B)[(A - B)^2 + 3AB] = (C + D)[(C + D)^2 - 3CD]
1(1^2 + 3AB) = 7(7^2 - 3CD)
1 + 3AB = 343 - 21CD
Substitute 42 - CD for AB because of (3)
1 + 3(42 - CD) = 343 - 21CD
Therefore, CD = 12

Substituting this into (3) gives :
AB = 42 - 12 = 30

Substituting B + 1 for A into AB = 30 gives :
(B + 1)B = 30
B^2 + B - 30 = 0
(B - 5)(B + 6) = 0
Therefore, B = 5 or -6

Therefore, A = 6 or -5, respectively.

Likewise, from CD = 12 and C = 7 - D we get :
(7 - D)D = 12
or, D^2 - 7D + 12 = 0
or, (D - 3)(D - 4) = 0
so, D = 3 or 4
Thus, C = 4 or 3, respectively.

Therefore, the solution sets for (A, B, C, D) are :
(6, 5, 3, 4) or (6, 5, 4, 3) or (-5, -6, 3, 4) or (-5, -6, 4, 3)

Answer 2

There are four solution sets, but the simplest is 6, 5, 4, 3. Let's check them out:

6 + 4 + 3 = 5 + 8
6² + 5² = 4² + 3² + 36
6*5 + 4*3 = 42
6³ = 5³ + 4³ + 3³