Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.
I remember I knew how to solve this at one point, but forgot over the summer.
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Suppose that x and y are two real numbers such that x-y=2 and x^2+y^2=8. Find x^3-y^3.
I remember I knew how to solve this at one point, but forgot over the summer.
x-y=2, now square both sides to get:Quote:
Originally Posted by ceasar_19134
(x-y)^2=x^2+y^2-2xy=8-2xy=2^2
so xy=2.
Now multiply x-y and x^2+y^2 together to get:
(x-y)(x^2+y^2)=x^3-y^3+xy^2-yx^2=(x^3-y^3) + xy(y-x)
But the the term at the extreme left is equal to 2.8=16, and the term at the
extreme right is (x^3-y^3)-4, hence x^3-y^3=20.
RonL
Please check the algebra carefully to make sure it's correct.
Hi,Quote:
Originally Posted by ceasar_19134
compare (x-y)^2 = x^2 - 2xy + y^2 = 4 and x^2 + y^2 = 8, thus 2xy = 4, thus xy = 2.
From x-y = 2 you get: x = y+2, therefore xy = 2 becomes: (y+2)y = 2
Simplify to: y^2 + 2y - 2 = 0
I leave the next steps to you. Substitute y in the equation x - y = 2 to calculate x.
I've got as final result to your problem x^3 - y^3 = 20.
tschüss
EB
Hello, ceasar_19134!
A variation of the Captain's solution . . . and Earboth's.
Quote:
Suppose that x and y are two real numbers such that:
. . [1] x - y .= .2 .and .[2] x² + y² .= .8
Find: x³ - y³
Square [1]: .(x - y)² .= .2² . → . x² - 2xy + y² .= .4 . → . (x² + y²) - 2xy .= .4
. . From [2] we have: .8 - 2xy .= .4 . → . -2xy = -4 . → . xy = 2 .[3]
Cube [1]: . (x - y)³ .= .2³ . → . x³ - 3x²y + 3xy² - y³ .= .8
. . We have: . x³ - y³ - 3xy(x - y) .= .8
From [2] and [3], we have: . x³ - y³ - 3(2)(2) .= .8
Therefore: . x³ - y³ .= .20
whats the next step after y^2+2y-2=o
Hi, that's a quadratic equation. Use the formula, which you probably know, to solve this equation:Quote:
Originally Posted by Rimas
If the equation is: ax^2 + bx + c = 0 then the solution is: x = 1/(2a)(-b+- sqrt(b^2-4ac))
Use this formula: y= 1/(2*1)(-2+- sqrt(2^2-4*1*(-2)) = -1+- sqrt(3)
tschüss
EB