if any one have guts then solve this:y^3=x^3+8x^2-6x+8?

Question

x,y are +ve integers
u have bto prove this also that it contain only limited roots

sorry it is "to" not " bto"

sorry it is "to" not " bto"

Answer 1

8x^2 - 6x + 8 is always positive for all x, since b^2-4ac is negative
(-220). (When the discriminant of a quadratic (b^2-4ac) is negative there are complex solutions only, i.e., no real roots, so the parabola defined by the equation is either always above or always below the x axis for all x, and this one is always above. Substituting one value shows this.)

Therefore y^3>x^3 and thus y>x
Let y = x+a where a is a positive integer.
Then
(x+a)^3 = x^3+8x^2-6x+8
x^3 + 3ax^2 + 3a^2x + a^3 = x^3 + 8x^2 - 6x + 8
or
(3a - 8)x^2 + (3a^2 + 6)x + (a^3 - 8) = 0
If a>2, then 3a-8, 3a^2+6, and a^3-8 are all positive, (as well as x^2 and x), therefore no solutions.

If a = 2 then,
-2x^2 + 18x = 0
or -2x(x-9) = 0
x = 9 (y=x+2=11) is the only positive solution.

a=1 yields
-5x^2 +9x - 7 = 0, and
no integral solutions (in fact no real ones)

whoever voted thumbs down should either re-read this or reevaluate themselves...
and I agree with Hal that rudeness is out of line. popcorn this means you - not to mention that you are wrong - maybe someone should make fun of you?

Answer 2

Some of you folks don't have the right to be so rude. Try being more polite. This is a good question.

Notice that

(x + 8/3)^3 = x^3+8x^2+(64/3)x+512/27,

So

y^3 = (x + 8/3)^3 - (82/3)x - 296/27
= (x + 8/3)^3 * (1 - ((82/3)x-296/27)/(x+8/3)^3 )

The latter part gets small for big x, so for increasing large positive x, y approaches

y ~ (x + 8/3), or x + 2.667.

The lesson here is that y is going to stay pretty close to x. We can therefore just consider all cases like

y = x + 1, or
y = x + 2, or
y = x + 3,

etc., because there won't be many to consider (y and x, as integers, can only differ by an integer, and it is easy to show (which you should do) that if x and y are both non-negative, then y > x.)

Let's compare each of these to our equation.

(x+1)^3 = x^3+3x^2+3x+1
(x+2)^3 = x^3+6x^2+12x+8
(x+3)^3 = x^3+9x^2+27x+27

It's easy to show (which you should do) that this last one is always bigger than
x^3+8x^2-6x+8

for non-negative x, and in general that

(x+h)^3 = x^3+3hx^2+3h^2x+h^3

is always bigger than

x^3+8x^2-6x+8

when h is an integer 3 or bigger. So these can never be equal - we won't get answers where y exceeds x by more than 2.

So this leaves the only possibilities as

y = x+1
y = x+2

Plug these into the original equation, cancel terms, and you get a quadratic equation in x for each. The y=x+1 case produces non-real roots for x; the y=x+2 case produces 0 and 9 for x.

So, the only possible solutions are
x=0, y=2
x=9, y=11

Answer 3

even as x is -a million the answer is 15 even as x is 0 that's 8 even as x is a million that's 3 even as x is two that's 0 even as x is 3 that's -a million even as x is 4 that's 0 even as x is 5 that's 3 even as x is 6 that's 8 and for the 2d question, contained in the first equation x=6-y and replace this contained in the first equation which provide you with 6-y=12+y so 2y=-6 and y= -3 . understanding y= -3 from first equation you'll locate x actual as x + (-3)=6 so x=9

Answer 4

Ok, I understand the nature of the question. There is only one solution: x=9, y=11 (note 0 is not a positive number)

The nature of the curve is such that at y-x = 2 when x=0, reaches a minimum when dy/dx=0, curves and again equals 2 when x=9, and asymptotically approaches 2.666... as x -> infinity. So there is only one answer. Not sure how to prove it.

Answer 5

I'm not sure how to prove it contains limited roots since the equation has an infinite range of values in quadrant 1. However using a little C# I found it has a solution where x = 9 and y = 11.

Answer 6

y^3 = x^3 + 8x^2 - 6x + 8
@x = 0, y = 2
@x = 9, y = 11

Answer 7

This eqn cant be solved .As per u got 2 variables and 1 eq.U need atleast 2 eqns to solve the mentioned problems.

Answer 8

y=cube root (x^3+8x^2-6x+8)

What are you asking for? A possible solution? (0,2)
x and y are positive even integers.

Answer 9

Hm, well, (0, 2) works...but 0's not positive.... I take it there's another one?

Answer 10

Two variables .... we need two equations to solve.......