Suppose that x and y are two real numbers such that x - y = 2 and x^2 + y^2 = 8.?

Question

Extremely hard

Find x^3 - y^3.

Answer 1

So far none has asnwered it fully

x-y = 2 ....1
x^2+y^2 = 8
we know (x+y)^2 + (x-y)^2 = 2(x^2+y^2)
put the values
(x+y)^2 + 4 = 16
(x+y)^2 = 12 ... 2
(x+y) = 2sqrt(3) ....3 or -2 sqrt(3)......4
from 1 and 3
by adding and dividing by 2
x = 1+sqrt(3)
so y = sqrt(3) -1
from 1 and 4
one can generate the 2nd set x= 1- sqrt(3) y = -1-sqrt(3)

now for (x^3-y^3) = (x-y)(x^2+y^2 + xy)
from 1 and 2 (x^2+y^2+xy) = 1/2((x+y)^2+x^2+y^2) = 1/2(8+12) = 10

so x^3-y^3 = 2* 10 = 20

Answer 2

If you change the first equation to x = y + 2 you can substitute it in the second equation. You'll get
(y + 2)^2 + y^2 = 8
y^2 + 4y + 4 + y^2 = 8 Since it's quadratic you need a zero on one side
2y^2 + 4y - 4 = 0 Divide everything by 2 to get rid of the common factor
y^2 + 2y - 2 = 0
This polynomial won't factor, so you'll have to solve for y with quadratic formula. Once you do that and simplify your answer you'll have y = -1 +/- square root 3

Substitute that answer to find x.

Sorry I'm not finishing it. Sometimes this stuff is a pain to type in. I hope it's a good start, though.

Answer 3

x^3-y^3 = 8

solve these two equations x-y=2 and x^2 +y^2 = 8

Replace the value of x, which is y+2, in x^2+Y^2=8 and solve the quardratic equation.

you will get x as 0 and Y as -2

Answer 4

substitute x=2+y into the second equation and solve the quadratic. Get value for x and y.
and you will find that x^3-y^3 = 20

Answer 5

It isn't hard at all!
You can't solve for x and y, as there are an infinite number of possible values. Everybody except the person talking about the circel has so far been wrong.
Just remember: 2^3=8
&)

Answer 6

Get x = 2+y from the first relation, plug it back into the second and solve the resulting quadratic equation for y
then plug the y or ys in the first relation for x
Now take both and plug into
x^3 - y^3
or prove algebraically.

Answer 7

use
x^3 - y^3 = (x - y) (x^2 + y ^2 + xy)

and

2xy = (x^2 + y ^2) - (x - y)^2

Then

xy = (8 - 2^2)/2 = 2

x^3 - y^3 = 2 * (8 + 2) = 20.

Answer 8

Answer is 12.

X-Y=2, X^2+Y^2=8.
(X-Y)^2=X^2+2XY+Y^2. Therefore using the above we obtain
xy=2,and then x^3-Y^3=(x-y)(X^2+Y^2-XY)=2*(8-2)=12

Answer 9

actually its quite easy - just the intersection of a straight line,
y=x-2 and a circle of radius 2*sqroot(2), center at origin.

i get x=1(+or-)sqroot(3), y=-1(+or-)sqroot(3). then substitute into x^3-y^3

Answer 10

it really is not symmetric to x-axis, y-axis, and/or starting place. no matter if it really is only asking for that it really is totally no longer symmetric to any of those, in the different case only pay interest to the others (x= -17/2)