How about giving us the solution? We can start from there and get back to the question. It is just to learn the steps that could be used to solve the question. No be so?

Ymodulus:

stories work am guy

baba why u dey tackle me now.at least i get idea.

Stop solving cos D more u look,the less u see!

Gudmorning gurus

I CRACKED IT!!!!!! x=4.958 y=-2.868 z=-1.09. It took me quite a while to crack dis.

Jack Ryan:

I CRACKED IT!!!!!! x=4.958 y=-2.868 z=-1.09. It took me quite a while to crack dis.

pls tel us how u did it.hope u didnt use a software or calculator.

No I didn't use any software or calculator. I solved it using a combination of substitution method, elimination method, binomial theorem and simultaneous equation method among others. And then I played with the equations I got some before I got what I was looking for. The key to solving this problem is to reduce all that bullshit to one equation in one variable and solving for that variable. When you finally get the equation in one variable, its going to be a polynomial to the order of 3, so you get 3 roots. If you are sharp you will notice that the three roots you get are going to be the same for the other two polynomials in the other two variables, so the three roots are the three variables.

Wao, mathematico!

I salute u, yes u are near d answer, or let me say u got it. But b4 d competitions, cowbell says it wil provide d solutions, but til now no solution, we shuld look foward to d answer, x=5, y=-3, z=-1, I salute Nairaland mathematicians, but i wil continue to solve it my self, thanx

Jack Ryan:

No I didn't use any software or calculator. I solved it using a combination of substitution method, elimination method, binomial theorem and simultaneous equation method among others. And then I played with the equations I got some before I got what I was looking for. The key to solving this problem is to reduce all that bullshit to one equation in one variable and solving for that variable. When you finally get the equation in one variable, its going to be a polynomial to the order of 3, so you get 3 roots. If you are sharp you will notice that the three roots you get are going to be the same for the other two polynomials in the other two variables, so the three roots are the three variables.

simply fantastic

This Solution is Amazing!!!

Indeed!

D; how did u solve it or u met a math prof.

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I found it useful and I don't want to be selfish; therefore, that's why I'm sharing this.

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its easy

there are 3 missing numbers

3x3=9

9x20=180

there are 180 degrees in a triangle

180x2=360

360 degrees is a circle

circle looks like an eye

OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG OMG

ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED ILLUMINATI CONFIRMED

have solve it , but the questions is like this, x + y + z= 1 x^2 + y^2 + z^2=35 , x^3 + y^3 + z^3 =97 .

Rasulullah:
Hey guys this question was given in question 1,essay part, 2009 FINAL COWBELL NASSMAC COMPETITION LAGOS, please we seriously need help on it.(the numbers after the letters are powers)
x+y+z=1
x2+y2+z2=34
x3+y3+z3=97

here we go
x + y + z = 1
x + y = 1 - z
x^2 + y^2 + z^2 = 34
x^2 + y^2 = 34 - z^2
(x + y)^2 - 2xy = 34 - z^2
from (1)
(1 - z)^2 - 2xy = 34 - z^2
(1 - 2z + z^2) - 2xy = 34 - z^2
2z^2 - 2z - 33 = 2xy
xy = (2z^2 - 2z - 33)/2
from 3
x^3 + y^3 + z^3 = 97
(x+y)^3 - 3x^2y - 3xy^2 = 97 - z^3
(x+y)^3 - 3xy(x+y) = 97 - z^3
I believe you can continue from here, all you do is substitute.

have already solve it , i got x=-3 or 5 ,y=5 or -3 and z= -1 twice .

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tobiope:
have solve it , but the questions is like this, x + y + z= 1 x^2 + y^2 + z^2=35 , x^3 + y^3 + z^3 =97 .

Very cheap question. I can play with it.

I can't believe I posted this question 7 years ago under my former username.

I can now clearly see how cheap this question is.
Refer to Advanced Mathematics books such as Backhouse & HK Dass. They have amazing answers to this. Thank you.