Question 387196: the product of three consecutive positive intergers is 8 times their sum. What is the sum of their squares? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! the product of three consecutive positive integers is 8 times their sum.
What is the sum of their squares?
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1st: x-1
2nd: x
3rd: x+1
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Equation:
x(x^2-1) = 8(3x)
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x^3-x = 24x
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x^3-25x = 0
x(x^2-25) = 0
x(x-5)(x+5) = 0
x = 0
x = 5
x = -5
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Using x = 5
1st: -4
2nd: 5
3rd: 6
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Sum of squares: (-4)^2+5^2+6^2 = 16+25+36 = 77
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Cheers,
Stan H.
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