B and J switch because J is a repeating number and repeating numbers are irrational because they don't have a definte end. Other than that, you are good!

Step-by-step explanation:

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2 years ago

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Use the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx) to determine the value of the sum of three integers given: the sum of

fenix001 [56]

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

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

the sum of their squares is 110, So $x^2+y^2 + z^2= 110$

the sum of their cubes is 684, so $x^3+y^3 + z^3= 684$

the product of the three integers is 210, so xyz= 210

the sum of any two products (xy+yz+zx) is 107

Now we plug in all the values in the identity

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

684 - 3(210) = (x+y+z)(110-107)

684 - 630 = (x+y+z)(3)

54 = 3(x+y+z)

Divide by 3 on both sides

18 = x+y+z

the value of the sum of three integers is 18

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3 years ago