If the square of the sum of three positive co | vedclass

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(D) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the square of their sum exceeds the sum of their squ

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$8$

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(D) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the square of their sum exceeds the sum of their squares by $292$.The sum of the numbers is $(x-1) + x + (x+1) = 3x$.The square of the sum is $(3x)^2 = 9x^2$.The sum of their squares is $(x-1)^2 + x^2 + (x+1)^2 = (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 3x^2 + 2$.Given that $9x^2 - (3x^2 + 2) = 292$.$6x^2 - 2 = 292$.$6x^2 = 294$.$x^2 = 49$.$x = 7$.The three numbers are $(7-1)$,$7$,and $(7+1)$,which are $6$,$7$,and $8$.The largest of the three numbers is $8$.

If the square of the sum of three positive consecutive natural numbers exceeds the sum of their squares by $292$,then what is the largest of the three numbers?

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