Question 1155188

Find three consecutive integers whose product is 161 larger than the cube of the smallest integer
<pre>Let the first integer be F
Then other 2 are: F + 1, and F + 2
We then get: {{{matrix(1,3, F(F + 1)(F + 2), "=", F^3 + 161)}}}
{{{matrix(1,3, F(F^2 + 3F + 2), "=", F^3 + 161)}}}
{{{matrix(1,3, 3F^2 + 2F - 161, "=", 0)}}}
{{{matrix(1,3, 3F^2 + 23F - 21F - 161, "=", 0)}}} ----- Substituting 23F - 21F for 2F
F(3F + 23) - 7(3F + 23) = 0 
F - 7 = 0               OR                    3F + 23 = 0
{{{highlight_green(matrix(2,4, First_number, "(F)", "=", 7,
"Others:", 8, "and", 9))}}}          OR          3F = - 23 (ignore)