Question

x+y+z=6
x²+y²+z²=26
x³+y³+z³= 90
Then find x⁴+y⁴+z⁴.​

Answer 1

36=(x+y+z)

2

=(x

2

+y

2

+z

2

)+2(xy+yz+zx)=26+2(xy+yz+zx)

⇒xy+yz+zx=(36−26)/2=10/2=5⇒xy+yz+zx=(36−26)/2=10/2=5

Using formula

(x+y+z)(x^2+y^2+z^2-xy−yz−zx)=x^3+y^3+z^3−3xyz(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)=x

3

+y

3

+z

3

−3xyz

We can find xyz:

6(26−5)=90−3xyz,126=90−3xyz,xyz=(90−126)/3=−36/3=−126(26−5)=90−3xyz,126=90−3xyz,xyz=(90−126)/3=−36/3=−12

Using formula x^4+y^4+z^4=(x+y+z)^4−4(x+y+z)^2(xy+yz+zx) +2(xy+yz+zx) ^2 +4(x+y+z)xyzx

4

+y

4

+z

4

=(x+y+z)

4

−4(x+y+z)

2

(xy+yz+zx)+2(xy+yz+zx)

2

+4(x+y+z)xyz

We can find x ^4 +y ^4 +z ^4 =6 ^4 −4∗6 ^2 ∗5+2∗5 ^2 +4∗6∗(−12)=338x

4

+y

4

+z

4

=6

4

−4∗6

2

∗5+2∗5

2

+4∗6∗(−12)=338

Answer: xyz=−12,x ^4 +y ^4 +z ^4 =338xyz=−12,x

4

+y

4

+z

4

=338