The product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?
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let x-1,x,x+1 are 3 consecutive positive numbers
given
product = 8(sum)
( x -1)x(x+1) =8(3x)
x(x^2-1) =24x
x can not 0
x^2-1=24
x^2=25
x=5
the numbers 4,5,6
sum of squares
16+25+36=77
given
product = 8(sum)
( x -1)x(x+1) =8(3x)
x(x^2-1) =24x
x can not 0
x^2-1=24
x^2=25
x=5
the numbers 4,5,6
sum of squares
16+25+36=77
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