Let a and b be real numbers such that a-b=4 and a^3 - b^3 = 52

(a) Find all possible values of ab

(b) Find all possible values of a+b

(c) Find all possible values of a and b

Guest Jan 28, 2023

#1**-1 **

(a)

By difference of cubes, a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Substituting and manipulating, we have a^3 - b^3 = 4[(a - b)^2 + 3ab]

Substituting again, we have a^3 - b^3 = 4(16 + 3ab) = 64 + 12ab = 52

Subtracting and dividing by 12, we have 12ab = 52 - 64 = -12, ab = -1.

Hence, **ab = -1**

proyaop Jan 28, 2023

#2**0 **

(b)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Earlier, we manipulated a^2 + ab + b^2 into (a - b)^2 + 3ab, however you can also manipulate it to (a + b)^2 - ab (do the expansion and see that it works!)

So... a^3 - b^3 = (a - b)[(a + b)^2 - ab] = 52

However, from (a), we learned that a - b = 4, and that ab = -1, substituting in, we have a^3 - b^3 = 4[(a + b)^2 + 1] = 52.

Dividing by 4, we have (a + b)^2 + 1 = 13, a + b = positive/negative square root of (12), which sqrt(12) = 2sqrt(3), so positive negative 2sqrt(3).

Hence, \(a + b = \pm2\sqrt{3}\)

proyaop Jan 28, 2023

#3**-1 **

(c)

Now that we know a + b = p/n 2sqrt(3), and that a - b = 4, we can use system of equations to get the individual values of each variable.

__Case 1:__ a + b = 2sqrt(3)

a + b = 2sqrt(3)

a - b = 4

Adding both equations, 2a = 4 + 2sqrt(3), a = 2 + sqrt(3).

Plugging in 2 + sqrt(3) for the top equation, 2 + sqrt(3) - 2sqrt(3) = -b, so b = sqrt(3) - 2.

__So for case 1__, \(a = 2 + \sqrt{3}\), and \(b = -2 + \sqrt{3}\)

__Case 2:__ a + b = -2sqrt(3)

a + b = -2sqrt(3)

a - b = 4

Adding both equations, 2a = 4 - 2sqrt(3), a = 2 - sqrt(3)

Plugging in 2 - sqrt(3) for the top equation, 2 - sqrt(3) + 2sqrt(3) = -b, so b = -2 - sqrt(3).

__So for case 2__, \(a = 2 - \sqrt{3}\), and \(b = -2 - \sqrt{3}\)

Hence, we have:

a = 2 + sqrt(3) => b = -2 + sqrt(3)

or

a = 2 - sqrt(3) => b = -2 - sqrt(3)

proyaop Jan 28, 2023

#5**0 **

Because each variable has a positive or negative value, we have to account for both possible values, creating two "cases", or options, and we test each one to get a different solution/ordered pair. If you don't understand, you should try searching "casework" problems, but this problem is just getting used to algebraic expansion. Remember difference of cubes!!!

proyaop
Jan 31, 2023