Algebra 2: Help PLZ (The question is pinned on)

Answers

Answer 1

Answer:

We know that:

$x^2+y^2+z^2=110\n\nx^3+y^3+z^3=684\n\nxyz=210\n\nxy+yz+zx=107$

so:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\n\n684-3\cdot210=(x+y+z)\big(110-(xy+yz+zx)\big)\n\n684-630=(x+y+z)(110-107)\n\n54=(x+y+z)\cdot3\qquad|:3\n\n\boxed{x+y+z=18}$

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Answers

29 is answer
Ok hope this helps

36-7=29

Hope this helps

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