# Determine the value of (1 + xyz)

• November 1st 2009, 10:52 PM
zorro
Determine the value of (1 + xyz)
Question : If x, y, z are all different and given that
$
\begin{vmatrix}
x & x^2 & 1 + x^3 \\
y & y^2 & 1 + y^3 \\
z & z^2 & 1 + z^3
\end{vmatrix} = 0,
$

Determine the value of (1 + xyz)
• November 1st 2009, 11:09 PM
Jodles
What have you done, and where are you stuck?
• November 1st 2009, 11:16 PM
math2009
$\det\begin{bmatrix} x&x^2&1+x^3\\ y&y^2&1+y^3\\ z&z^2&1+z^3 \end{bmatrix}=\cdots=(z-x)(z-y)(y-x)(1+xyz)=0$, if you have enough patience.

$\because x\neq y\neq z\neq x \ \therefore 1+xyz=0$
• November 2nd 2009, 02:50 PM
mr fantastic
• December 1st 2009, 12:29 PM
zorro
Could u please provide the end result
Quote:

Originally Posted by mr fantastic

I have seen the reply post in the other thread but cold u please provide me with the end answer, just to be sure that i have done it right or no.At present i am trying to do as provided by the prev thread but it would be good if u could provide with the end result
ie, $(1 + xyz)$
• December 2nd 2009, 12:38 AM
mr fantastic
Quote:

Originally Posted by zorro
I have seen the reply post in the other thread but cold u please provide me with the end answer, just to be sure that i have done it right or no.At present i am trying to do as provided by the prev thread but it would be good if u could provide with the end result
ie, $(1 + xyz)$

You need to start showing some effort here. Post all your work. Say where you get stuck.
• December 22nd 2009, 04:45 PM
zorro
Is this correct?
Quote:

Originally Posted by mr fantastic
You need to start showing some effort here. Post all your work. Say where you get stuck.

After doing the determinant i am getting the following please advice if correct or no

$(xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2)(1+xyz)=0$

$\therefore (1+xyz) = 0$
• December 23rd 2009, 01:15 AM
mr fantastic
Quote:

Originally Posted by zorro
After doing the determinant i am getting the following please advice if correct or no

$(xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2)(1+xyz)=0$

$\therefore (1+xyz) = 0$

• December 23rd 2009, 01:17 AM
zorro
Quote:

Originally Posted by mr fantastic

Just wanted to check i did it correctly or no .........want to know if the step is correct or no?????
• December 23rd 2009, 01:27 AM
mr fantastic
Quote:

Originally Posted by zorro
Just wanted to check i did it correctly or no .........want to know if the step is correct or no?????

Look at post #3 and compare answers. Does $(xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2)$ (which is what you have) equal $(z-x)(z-y)(y-x)$ ? (the answer in post #3 is correct, by the way).
• December 25th 2009, 06:23 PM
simplependulum
Quote:

Originally Posted by zorro
After doing the determinant i am getting the following please advice if correct or no

$(xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2)(1+xyz)=0$

$\therefore (1+xyz) = 0$

I think some marks will be deducted in your solution .

You should mention that iff two of the elements are the same , the factor

$xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2 = 0$

because you have only showed that

$xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2 = 0$

OR

$1 + xyz = 0$

so $1 + xyz$ is not 100% zero

Factorising it you will get full mark ! (Happy)