₦airaland Forum

PhilAmadeus:
Omo dis animals are cute oooo...I knw say nah that Aye Aye own go dey different see hin Eye lyk Halogen bulb,love d animal look funny to me

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.

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

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

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.

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.

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

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.

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

udemzy101:
Still prefer airtel 1.5gb for #200


At night I do get up to 800 km /sec - 1mb/sec


That's enough, if them like make them stay there

DollarAngel:
DON BABA J, HAS DONE WHAT NOISE MAKING OLAMIDE CAN'T OR WON'T DO, LOUSY ARTIST, THIS IS HOW TO GIVE BACK TO THE SOCIETY

OTAIFOHJ:
Remember that WIKI simply means "What I Know Is"

Meanwhile,


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stanway:
I dialled *449*1# several times before I finally got mine just keep dialling

FabioPeter:
Some idiots are taking shots.

generaldre:
Lol Can you beat this?

OGeeKush:
Baba hwfa na i see say na handler things u dey use.. Epp ur boy na

ToriBlue:
You mean glo browse free in ucmini?

ToriBlue:
Blood of......... Please share me some data

ugotex85:
Thanks man, but Is it still working?

zekokuyo:
For 2 weeks

ugotex85:
My guy help us with the cheats, make we follow you enjoy life.