Question

I have few problems:
1) x+y+z=0
x^2+y^2+z^2=2
x^4+y^4+z^4=?

2) x+y+z=a
x^2+y^2+z^2=3a^2/2
x^3+y^3+z^3=a^3
xy+yz+xz=?
xyz=?
x^4+y^4+z^4=?

3) x/a + y/b + z/c = 1
a/x + b/y + c/z = 0
x^2/a^2 + y^2/b^2 + z^2/c^2 = ?

4) (7+5*(2^(1/2)))^(1/3)+(7-5*(2^(1/2)))^(1/3)

Answer 1

4) \[\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\]

Answer 2

R u there? If so let's look at each problem.

Answer 3

i am

Answer 4

What is the topic that you got these questions from?

Answer 5

powers and roots

Answer 6

What does the directions for these questions say?

Answer 7

no directions. just what I wrote here. in first 3 questions, I have some expressions given and I have to evaluate those with question mark. The 4th one I just have to calculate. I know that I can use Lagrange's identity for sqrt, but I don't know what to do with 3th root.

Answer 8

Sorry wish I could help. This is the 1st time I have come across these type of questions. I will keep looking into them if I am able to find the solutions I will try to post it, if you are still stuck.

Answer 9

sure, if u find solution - post it :)

Answer 10

I found the solution for 1 and 2:
#1: x=-1,y=1,z=0 or change it around as u please.
#2: x=-a/2, y=a/2 and z=a or change it around as u see fit.

Answer 11

Put these solution in the equation you are trying to find and solve from there.
#1: Answer=2

Answer 12

but I have to write how did I get it o.O

Answer 13

Since, x^2+y^2+z^2=2
-1<=x<=1 , -1<=y<=1 and -1<=z<=1

Answer 14

From, x+y+z=0
u will get one of them is -1 , the other one is 1 and the other must be 0

Answer 15

So, x^4+y^4+z^4=2

Answer 16

\[\large xy+yz+xz=-\frac{ 1 }{ 4 }a^2\]\[\large xyz=-\frac{1}{4}a^3\]using algebra

Answer 17

3) x/a + y/b + z/c = 1
a/x + b/y + c/z = 0
x^2/a^2 + y^2/b^2 + z^2/c^2 =1

Answer 18

Let x/a=i , y/b=j , z/c=k
then,
i+j+k=1

1/i + 1/j+1/k=0
jk+ik+ij=0

Now,
(i+j+k)^2=i^2+j^2+k^2+2(ij+ik+jk)
(1)^2=i^2+j^2+k^2+2*0
Thus, i^2+j^2+k^2=1
x^2/a^2 + y^2/b^2 + z^2/c^2 =1

Answer 19

LOL this looks familiar. For the first one, except for saurav's logical reasoning, you can go for this:
\((x+y+z)^2=0\)
\(x^2+y^2+z^2 + 2(xy+xz+yz)=0\)
\(2 + 2(xy+xz+yz)=0\)
\(xy+xz+yz=-1\)
\((x^2+y^2+z^2)^2=x^4+y^4+z^4 + 2(x^2y^2+x^2z^2+z^2y^2)\)
\((xy+xz+yz)^2=1\)
\(x^2y^2+x^2z^2+z^2y^2 + 2(xyz)(x+y+z)=1\)
\(x^2y^2+x^2z^2+z^2y^2 =1\)
so,\((2)^2=x^4+y^4+z^4 + 2(1)\)
\(x^4+y^4+z^4 =2\)

Answer 20

thank you, guys :)

Answer 21

You're welcome :)