If the sum of the cubes of two numbers is 72 and the difference of their squres is 12. Find the numbers.
x^3 + y^3 = 72
x^2 - y^2 = 12
We could just do this by listing the cubes less than 72 :
1, 8, 27, 64
the only combination for a sum of 72 is 8 + 64
so x=2, y=4
so the numbers are 2 and 4
is the difference property ok?
is 16-4 = 12 ?? YES
also, we have
(x+y)(x-y) = 12
The pairs of factors of 12 are
1,12 2,6 3,4
x+y=6
x-y=2
x=4,y=2
To solve this problem, let's call the two numbers "x" and "y". We are given two conditions:
1. The sum of the cubes of the two numbers is 72, which we can express as:
x^3 + y^3 = 72
2. The difference of their squares is 12, which can be written as:
x^2 - y^2 = 12
Now, let's use these two equations to find the values of x and y.
First, we can simplify the second equation by factoring it as a difference of squares:
(x - y)(x + y) = 12
Since we are looking for the values of x and y, we need to isolate one of the variables in one of the equations, and substitute it into the other equation. Let's solve the second equation for x - y:
x - y = 12 / (x + y)
Now, we can substitute this value into the first equation:
(x - y)(x^2 + xy + y^2) = 72
Substituting x - y = 12 / (x + y):
(12 / (x + y))(x^2 + xy + y^2) = 72
Now, let's simplify this equation:
12(x^2 + xy + y^2) = 72(x + y)
Divide both sides by 12 to further simplify the equation:
x^2 + xy + y^2 = 6(x + y)
Now, let's work on simplifying this equation further. Rearrange the equation:
x^2 + xy + y^2 - 6x - 6y = 0
Now, let's focus on simplifying the left-hand side of the equation. We can complete the square for the variables x and y by adding and subtracting appropriate terms:
(x^2 - 6x + 9) + (y^2 - 6y + 9) - 18 = 0
(x - 3)^2 + (y - 3)^2 - 18 = 0
Now, let's simplify this equation further:
(x - 3)^2 + (y - 3)^2 = 18
From this equation, we can see that the sum of the squares of the differences between x and 3, and y and 3 is equal to 18.
At this point, we can notice that there are two possible pairs of numbers that satisfy this equation. They are (3, 3) and (-3, -3). These two pairs of numbers have cubes that sum up to 72 and squares that differ by 12.
Therefore, the numbers that satisfy both conditions are x = 3 and y = 3, or x = -3 and y = -3.