If x + y + z = 2, x^2 + y^2 + z^2 = 6, and x^3 + y^3 + z^3 = 8, what i

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If  x + y + z = 2 ,  x^2 + y^2 + z^2 = 6 , and  x^3 + y^3 + z^3 = 8 , what is the value of  x^4 + y^4 + z^4  ?

A. 16
B. 8√5
C. 18 
D. 18√3 
E. 30

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GMATinsight · Accepted Answer

My first attempt was to take 30 seconds and think of the numbers that could satisfy the two expressions

x+y+z=2 
and 
 x^2+y^2+z^2=6

Since six is square of three numbers  1^2+1^2+2^2 
but then sum of these three numbers was to be brought 2 hence it must be  (-1)^2+1^2+2^2  so that  (-1)+1+2=2

Also for these values  x3+y3+z3=(-1)^3+1^3+2^3=8

therefore  x^4+y^4+z^4 = (-1)^4+1^4+2^4 = 1+1+16 = 18

Answer: Option C

vg18 · Answer

just tried with some numbers:
x=-1
y=1
z=2
satisfies all the equations. hence ans is 18,C.

freedom128 · Answer

This is difficult to solve under 5 mins. Is it really gmat like?

Let
a= x+y+z = 2
b= x^2+y^2+z^2 = 6
c= x^3+y^3+z^3 = 8

* d= xy+yz+xz = (a^2-b)/2 = (2^2-6)/2 = -1
* e= xyz = (a^3-3ab+2c)/6 = (2^3-3.2.6+2.8)/6 = -2

* x^4+y^4+z^4
= (x^2+y^2+z^2)^2 - 2{x^2.y^2+y^2.z^2+x^2.z^2}
= (x^2+y^2+z^2)^2 - 2{(xy+yz+xz)^2 - 2xyz(x+y+z)}
= b^2 - 2{d^2-2e.a}
= 36 - 2{1-2(-2).2}
= 36 - 2.9
= 18

FINAL ANSWER IS (C)

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unraveled · Answer

If x+y+z=2,  x^2+y^2+z^2=6 , and  x^3+y^3+z^3=8 , what is the value of  x^4+y^4+z^4  ?

A. 16
B. 8√5
C. 18
D. 18√3
E. 30

Some important formula first. 
 (x + y + z)^2 = x^2+y^2+z^2 + 2(xy + yz + zx)  - Eqn. 1
 x^4+y^4+z^4 = (x^2+y^2+z^2)^2−2x^2y^2−2y^2z^2−2x^2z^2  - Eqn. 2
 x^3+y^3+z^3−3xyz=(x+y+z)(x^2+y^2+z^2−xy−yz−xz)=(x+y+z)^3−3(xy+yz+xz)(x+y+z)  - Eqn. 3

From Eqn. 1
2(xy + yz + zx) = 2^2 - 6
(xy + yz + zx) = -1

From Eqn. 3
8 - 3xyz = 2^3 - 3(-1).(2)
xyz = -2

Also  (xy + yz + zx)^2 = x^2y^2+y^2z^2+z^2x^2 + 2xyz(x + y + z) 
So  x^2y^2+y^2z^2+z^2x^2 = 1 - 2(-2)(2) = 9

Hence from Eqn. 2
 x^4+y^4+z^4 = (x^2+y^2+z^2)^2−2x^2y^2−2y^2z^2−2x^2z^2 
 x^4+y^4+z^4 = 6^2−2.9 = 18

Answer C.

lacktutor · Answer

If  x+y+z=2 ,  x^{2}+y^{2}+z^{2}=6 , and  x^3+y^3+z^3=8 , what is the value of  x^4+y^4+z^4=  ?

Since nothing tells us about whether x,y,z -integer, positive, negative number, Let's use number picking approach to get to the answer.

-->  x=1 ,  y=-1  and  z=2  
 1+(-1)+2=2

1^{2}+(-1)^{2}+2^{2}=6

1^{3}+(-1)^{3}+2^{3}=8

-->  1^{4}+(-1)^{4}+2^{4}=1+1+16 =18

The answer is C.

eakabuah · Answer

Given that x+y+z=2 ------(1), x^2+y^2+z^2=6 ------(2), and x^3+y^3+z^3=8 -------(3), we are to determine x^4+y^4+z^4
Now, looking at (1), (2) and (3), x, y, and z must be integers because there is no way we can get x+y+z=2 and x^2+y^2+z^2=6 if any of the values of x,y, or z is not an integer.
Secondly, one of the variables must be 2, so lets say y=2, and the other variables must sum to zero. The only values of x and z that sum to zero that satisfy (1) and (2) are x=1 and z=-1 or vice versa.
since x and z are negative reciprocals, x+z=0 and x^3+z^3=0.
Hence x+y+z=y=2 and x^3+y^3+z^3=y^3=2^3=8.
So, x^4+y^4+z^4=1^4+2^4+(-1)^4=16+2=18

The answer is therefore C.

GMATWhizTeam · Answer

Solution:  
The best way to solve this kind of problem is hit and trial method. We will plug in the values and see if they satisfy the given equations.
 • We will choose small integral values of  x, y,  and  z  so that finding square and cubes will be easier.
• We are given that  x +y + z = 2.  Let’s choose any two integers to be  0 , in that case, we will have only one integer to solve.
 o If  x = y = 0 , then  x + 0 + 0 = 2 
   x=2, y=0 , and  z=0  
o Now, substitute these values in the second equation
   x^2+y^2+z^2 = 2^2 + 0 + 0 = 4 , these values does not satisfy the second equation. 
o Now, think of different values, If  y =1 , and  z =-1 , then  x +1 -1 =2 
   x =2, y= 1, z =-1  
o Now, Substitute these values in the second equation.
   x^2+y^2+z^2=2^2+1^2+(-1)^2 =4 +1+1=6 , satisfied. 
o Next, we will substitute these values in the third equation.
   x^3+y^3+z^3=2^3+1^3+(-1)^3=8+1-1=8 , satisfied. 
o Thus, we can find the value of  x^4+y^4+z^4 , with these values.
   2^4+1^4+(-1)^4=16+1+1=18 .  
 Hence, the correct answer is  Option C.

sujoykrdatta · Answer

x⁴+y⁴+z⁴
=(x²+y²+z²)² - 2(x²y² + y²z² + x²z²)
=(x²+y²+z²)² - 2

= (x²+y²+z²)² - 2

= (x²+y²+z²)² - 2  ... (i)

Now: 
(x+y+z)²=(x²+y²+z²)+2(xy+yz+zx)
=> 2²=6+2(xy+yz+zx)
=> xy+yz+xz = -1 ... (ii)

Again: 
x³+y³+z³-3xyz = (x+y+z)

=> 8 - 3xyz = 2 
=> xyz = -2 ... (iii)

Using (ii) and (iii) in (i):

x⁴ + y⁴ + z⁴ = (x²+y²+z²)² - 2

=> x⁴ + y⁴ + z⁴ = 6² - 2

=> x⁴ + y⁴ + z⁴ = 36 - 2  = 18

Answer C

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If x + y + z = 2, x^2 + y^2 + z^2 = 6, and x^3 + y^3 + z^3 = 8, what i

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