SOLUTION: if a,b,c are three numbers such that cube of first number exceeds the product of three numbers by 2, the product of three numbers exceeds the cube of second number by 3and the cube

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Question 623884: if a,b,c are three numbers such that cube of first number exceeds the product of three numbers by 2, the product of three numbers exceeds the cube of second number by 3and the cube of third number exceeds the product of three numbers by 3. find the numbers.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If a, b and c are the first, second and third numbers, respectively, and if "the product of three numbers" means the product of a, b and c, then
  • "cube of first number exceeds the product of three numbers by 2" translates into:
    a%5E3+=+a%2Ab%2Ac+%2B+2
  • "the product of three numbers exceeds the cube of second number by 3" translates into:
    a%2Ab%2Ac+=+b%5E3+%2B+3
  • "the cube of third number exceeds the product of three numbers by 3" translates into:
    c%5E3+=+a%2Ab%2Ac+%2B+3
We now have three equations with 3 variables. We should be able to solve this system. Fortunately every equation has the expression a*b*c in it. We can do a lot of simplifying if we start by solving each equation for a*b*c. From the first equation we get:
a%2Ab%2Ac+=+a%5E3-2
The second equation is already solved for a*b*c. From the third equation we get:
a%2Ab%2Ac+=+c%5E3+-+3

All these a*b*c's must be equal. So we can get "new" equations:
a%5E3-+2+=+b%5E3%2B3
a%5E3-+2+=+c%5E3-3
and
b%5E3%2B3+=+c%5E3-3

In these equations all the variables are cubes. To simplify matters I am going to temporarily replace them with other variables:
Let p+=+a%5E3, q+=+b%5E3 and r+=+c%5E3. Now our equations are:
p - 2 = q + 3
p - 2 = r - 3
q + 3 = r - 3

Now we solve this system. You have probably learned a number of methods for solving systems like this. Any of them should work. You do not have to solve it the following way.

Putting it into a "standard" form:
p - q     = 5
p     - r = -1
    q - r = -6

Subtracting the second equation from the first we get:
-q + r = 6
Adding this equation to the third equation we get:
0 = 0
The fact that both q and r disappeared at the same time tells us that there is not a single solution to this problem. The fact that the statement, 0 = 0, is true tells us that there are an infinite number of solutions to this problem.

If there is supposed to be a single solution (as the phrase "find the numbers" suggests), then there is either an error in what I've done, or there is an error in what you posted or there was an error in the problem you were given. I have double-checked my work and I have even tried an alternate method (Cramer's Rule) of solving the system of equations and I have not been able to find an error in my work.

Please re-post your problem after double-checking both the problem and how you post it.

P.S. If we had been able to find values for p, q and r then we would find their cube roots to find a, b and c.