Question

Given that real numbers x,y,z satisfy the system of equations
x+y+z=6
x^2+y^2+z^2=26
x^3+y^3+z^3=90
Find the values of x×y×z and x^4+y^4+z^4

Answer 1

$36=(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)=26+2(xy+y z+zx)$

$⇒xy+yz+zx=(36−26)/2=10/2=5$

Using formula

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

We can find xyz:

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

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

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

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