Pls guys, I've attached the question. I did it to a point and got my value of xyz to be 15. I just can't get the numbers themselves.

Not this morning... Run out of thread

X+y+z=1.....(1) X^2+y^2+z^2=35......(2) X^3+y^3+z^3=97.......(3) From equ(3) We that x^3+y^3=(x+y)^3-3xy(x+y) So input that into equation(3) (x+y)^3-3xy(x+y)+z^3=97....(4) Using equ(1) x+y=1-z....(5) Subt into equ(4) (1-z)^3-3xy(1-z)+z^3=97 1-3z+3z^2-z^3-3xy(1-z)+z^3=97 Collect like terms -3z+3z^2-3xy(1-z)=96......(6) From equ(2) x^2+y^2=35-z^2....(7) Also x^2+y^2=(x+y)^2-2xy.....( But x+y=1-z so x^2+y^2=(1-z)^2-2xy....(9) Equate equ(7) to equ(9) 35-z^2=(1-z)^2-2xy Make xy the subject and subt into equ(6) xy=z^2-z-17 Put it into -3z+3z^2-3xy(1-z)=96 -3z+3z^2-3(z^2-z-17)(1-z)=96 Divide through by 3 andExpand the bracket and collect like terms z^3-z^2-17z-15=0 Solve the polynomial z=-1, z=-3 z=5 Subt for x and y in each value of z x=-1, x=-3 x=5 And y=-1, y=-3, y=5.

RealLordZeus:
Not this morning... Run out of thread

haha. Abeg help bro. It's not compulsory you must solve it in the morning. You can solve it in the afternoon.

mikezuruki:
X+y+z=1.....(1) X^2+y^2+z^2=35......(2) X^3+y^3+z^3=97.......(3) From equ(3) We that x^3+y^3=(x+y)^3-3xy(x+y) So input that into equation(3) (x+y)^3-3xy(x+y)+z^3=97....(4) Using equ(1) x+y=1-z....(5) Subt into equ(4) (1-z)^3-3xy(1-z)+z^3=97 1-3z+3z^2-z^3-3xy(1-z)+z^3=97 Collect like terms -3z+3z^2-3xy(1-z)=96......(6) From equ(2) x^2+y^2=35-z^2....(7) Also x^2+y^2=(x+y)^2-2xy.....( But x+y=1-z so x^2+y^2=(1-z)^2-2xy....(9) Equate equ(7) to equ(9) 35-z^2=(1-z)^2-2xy Make xy the subject and subt into equ(6) xy=z^2-z-17 Put it into -3z+3z^2-3xy(1-z)=96 -3z+3z^2-3(z^2-z-17)(1-z)=96 Divide through by 3 andExpand the bracket and collect like terms z^3-z^2-17z-15=0 Solve the polynomial z=-1, z=-3 z=5 Subt for x and y in each value of z x=-1, x=-3 x=5 And y=-1, y=-3, y=5.

. Thanks very much bro. I'm very grateful.

X=5
Y= -3
Z= -1

nextprince:
X=5
Y= -3
Z= -1

Thanks very much bro. .

sureteeboy:

Pls, bro, can you help me snap your working sheet and upload it.
I learnt that the equation has 6 true answers (that is, 3 different but true values for x y and z). You've got one of them and Mikezuruki also got another one, so, I want to see how the method differs. Thanks very much.

In this equation, there is no unique value attached to any of the unknowns. They are interchangeable. Attempt interchanging the values I stated and check, u will arrive at the same answer.

Eg
5-3-1=1
-3+5-1=1
-1-3+5=1

Or

5^2-3^2-1^2=35
-3^2+5^2-1^2=35
-1^2-3^2+5^2=35
Same with the third equation.
Such elaborate working ain't necessary especially when time is a factor.
From equation 1, it is very apparent that there are negative and positive values
And from equation 2, the square of any number is positive and certainly non of the numbers should be more than 6. Therefore, u can choose from -4 to 5 and be adding the squares of different combinations.
I got it at equation 2, and simply confirm it with equation 3.

nextprince:


In this equation, there is no unique value attached to any of the unknowns. They are interchangeable. Attempt interchanging the values I stated and check, u will arrive at the same answer.

Eg
5-3-1=1
-3+5-1=1
-1-3+5=1

Or

5^2-3^2-1^2=35
-3^2+5^2-1^2=35
-1^2-3^2+5^2=35
Same with the third equation.
Such elaborate working ain't necessary especially when time is a factor.
From equation 1, it is very apparent that there are negative and positive values
And from equation 2, the square of any number is positive and certainly non of the numbers should be more than 6. Therefore, u can choose from -4 to 5 and be adding the squares of different combinations.
I got it at equation 2, and simply confirm it with equation 3.

Thanks Bro. I figured that out, that's why I deleted the question. I never knew you saw it.

(x, y, z) = (-1, -3, 5) OR (x, y, z) = (-3, 5, -1) OR (x, y, z) = (5, -1, -3)

mikezuruki:
X+y+z=1.....(1)
X^2+y^2+z^2=35......(2)
X^3+y^3+z^3=97.......(3)
From equ(3)
We that x^3+y^3=(x+y)^3-3xy(x+y)
So input that into equation(3)
(x+y)^3-3xy(x+y)+z^3=97....(4)
Using equ(1) x+y=1-z....(5)
Subt into equ(4)
(1-z)^3-3xy(1-z)+z^3=97
1-3z+3z^2-z^3-3xy(1-z)+z^3=97
Collect like terms
-3z+3z^2-3xy(1-z)=96......(6)
From equ(2)
x^2+y^2=35-z^2....(7)
Also x^2+y^2=(x+y)^2-2xy.....(
But x+y=1-z so
x^2+y^2=(1-z)^2-2xy....(9)
Equate equ(7) to equ(9)
35-z^2=(1-z)^2-2xy
Make xy the subject and subt into equ(6)
xy=z^2-z-17
Put it into -3z+3z^2-3xy(1-z)=96
-3z+3z^2-3(z^2-z-17)(1-z)=96
Divide through by 3 andExpand the bracket and
collect like terms
z^3-z^2-17z-15=0
Solve the polynomial
z=-1, z=-3 z=5
Subt for x and y in each value of z
x=-1, x=-3 x=5
And y=-1, y=-3, y=5.

Pls Bro, I solved the polynomial but I don't understand what you meamt by ;substitute for x and y in each value of z.

sureteeboy:

Pls Bro, I solved the polynomial but I don't understand what you meamt by ;substitute for x and y in each value of z.

If you've gotten the unknown(s) for x and y, no need solving for z. Just subtitute the figures gotten for both x and y

mikezuruki:

If you've gotten the unknown(s) for x and y, no need solving for z. Just subtitute the figures gotten for both x and y

No, what I meant was, I solved the polynomial and I got the 3 values of z. Now, how do I get the values of x and y from there?

mikezuruki:

If you've gotten the unknown(s) for x and y, no need solving for z. Just subtitute the figures gotten for both x and y

Bro please, kindly assist for this last time. After getting those values of z, I don't know where to substitute them to get each values of x and y

sureteeboy:
Bro please, kindly assist for this last time. After getting those values of z, I don't know where to substitute them to get each values of x and y

Then find x or y the same way you found z. You only need to know two of the unknowns to solve for x, y [/i]and [i]z.

Tellemall:


Then find x or y the same way you found z. You only need to know two of the unknowns to solve for x, y [/i]and [i]z.

Yeah. I've already done it. Thanks very much.