Question 1209229: The real numbers a and b satisfy a - b = 2 and a^3 - b^3 = 8.
(a) Find all possible values of ab.
(b) Find all possible values of a + b.
(c) Find all possible values of a and b.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
The real numbers a and b satisfy a - b = 2 and a^3 - b^3 = 8.
(a) Find all possible values of ab.
(b) Find all possible values of a + b.
(c) Find all possible values of a and b.
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Factor a^3 - b^3
a^3 - b^3 = (a-b)*(a^2 + ab + b^2).
Then from the second equation
(a-b)*(a^2 + ab + b^2) = 8.
Replace here a-b by 2, based on the first equation. You will get
a^2 + ab + b^2 = 4.
Substitute here b = a-2. You will get
a^2 + a*(a-2) + (a-2)^2 = 4,
a^2 + a^2 - 2a + a^2 - 4a + 4 = 4,
3a^2 - 6a = 0,
3a(a-2) = 0.
It gives two roots for "a" : a = 0 and a = 2.
If a= 0, then b = a-2 = -2.
If a= 2, then b = a-2 = 0.
So, the solutions for the given system of equations are these pairs (a,b) = (0,-2) and (a,b) = (2,0).
Having this, you can compute everything what you want / (you need) and answer all the questions.
Solved.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Part 1) Find all possible values of ab.
(a-b)^3
= (a-b)(a-b)^2
= (a-b)(a^2-2ab+b^2)
= a(a^2-2ab+b^2)-b(a^2-2ab+b^2)
= (a^3-2a^2b+ab^2)+(-a^2b+2ab^2-b^3)
= a^3 + (-2a^2b-a^2b) + (ab^2+2ab^2) - b^3
= a^3 - 3a^2b + 3ab^2 - b^3
= a^3-b^3-3ab(a-b)
In short,
(a-b)^3 = a^3-b^3-3ab(a-b)
You can skip over the previous paragraph if you have this formula memorized or written on a notecard.
Then we apply the equations a-b = 2 and a^3-b^3 = 8 to isolate ab.
So,
(a-b)^3 = a^3-b^3-3ab(a-b)
(a-b)^3 = a^3-b^3-3ab(a-b)
(2)^3 = 8-3ab(2)
8 = 8 - 6ab
-6ab = 8-8
-6ab = 0
ab = 0/(-6)
ab = 0
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Part 2) Find all possible values of a+b.
c = a+b
c^2 = a^2+2ab+b^2
c^2 = a^2+2*0+b^2 ..... plug in ab = 0
c^2 = a^2+b^2
Use the difference of cubes factoring formula
a^3-b^3 = (a-b)(a^2+ab+b^2)
a^3-b^3 = (a-b)(a^2+0+b^2)
a^3-b^3 = (a-b)(a^2+b^2)
a^3-b^3 = (a-b)c^2
8 = 2c^2
c^2 = 8/2
c^2 = 4
c = sqrt(4) or c = -sqrt(4)
c = 2 or c = -2
a+b = 2 or a+b = -2
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Part 3) Find all possible values of a and b.
a-b = 2
a = b+2
a^3 - b^3 = 8
(b+2)^3 - b^3 = 8
(b^3+3*b^2*2+3*b*2^2+2^3) - b^3 = 8
6b^2+12b+8 = 8
6b^2+12b = 0
6b(b+2) = 0
6b = 0 or b+2 = 0
b = 0 or b = -2
If b = 0, then a = b+2 = 0+2 = 2
One ordered pair solution is (a,b) = (2,0)
If b = -2, then a = b+2 = -2+2 = 0
The other ordered pair solution is (a,b) = (0,-2)
Note that you can do part 3 first to determine a,b
Then it's very easy to compute ab and a+b.
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Answers:
ab = 0
a+b = 2 or a+b = -2
(a,b) = (2,0) or (a,b) = (0,-2)
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