## gorica 2 years ago 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)

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1. gorica

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

2. philo1234

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

3. gorica

i am

4. philo1234

What is the topic that you got these questions from?

5. gorica

powers and roots

6. philo1234

What does the directions for these questions say?

7. gorica

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.

8. philo1234

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.

9. gorica

sure, if u find solution - post it :)

10. philo1234

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.

11. philo1234

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

12. gorica

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

13. sauravshakya

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

14. sauravshakya

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

15. sauravshakya

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

16. sirm3d

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

17. sauravshakya

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

18. sauravshakya

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

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,$$(2)^2=x^4+y^4+z^4 + 2(1)$$ $$x^4+y^4+z^4 =2$$

20. gorica

thank you, guys :)