Question

.

Answer 1

\[
z=x+y\implies z^2=x^2+2xy+y^2
\]

Answer 2

So \[
x^2+y+2x=x^2+2xy+y^2
\]

Answer 3

uhg, that was redundant.

Answer 4

if \[
y+2x=y(y+2x)
\]

Answer 5

then maybe \(y=1\)

Answer 6

Then \(z=x+1\)

Answer 7

\[
x^3+1=(x+1)^2-168
\]

Answer 8

It is either 1,2,3,4,5,6,7,8,9

Answer 9

Wait, I think your algebra is wrong.

Answer 10

\(x^3\) should cancel.

Answer 11

The \(^2\) should be \(^3\) if you went off what I wrote.

Answer 12

\[
x^3+1=(x+1)^3-168
\]

Answer 13

\[
(x^3+1) - (x+1)^3=-168
\]the first and last terms cancel out.

Answer 14

using pascals triangle we know our terms are
1
1 1
1 2 1
1 3 3 1

Answer 15

So \[
-3x^2-3x = -168
\]

Answer 16

this is a quadratic equation

Answer 17

I get \(x=7,y=1,z=8\)