I only know this because I remember the common fractions, and 1/3 is .3333 repeating.
In a barangay fiesta singing contest with 12 contestants , in how many ways can the judges choose the first 3 winners
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I only know this because I remember the common fractions, and 1/3 is .3333 repeating.
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the sum of their squares is 110, So
the sum of their cubes is 684, so
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
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