document.write( "Question 1155188: Find three consecutive integers whose product is 161 larger than the cube of the smallest integer \n" ); document.write( "

Algebra.Com's Answer #777738 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi  
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document.write( " Let n represent the smallest integer  
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document.write( "n(n+1)(n+2) = n^3- 161
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document.write( "n(n^2 + 3n + 2)= n^3- 161
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document.write( "n^3 + 3n^2 + 2n = n^3 + 161
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document.write( "3n^2 + 2n - 161 = 0
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document.write( "  Integer solution
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document.write( "Integers 7 , 8 , 9
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document.write( "CHECKING our answer***  7*8*9 = 7^3 + 161 = 504
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document.write( "Wish You the Best in your Studies.
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document.write( "Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation  (in our case ) has the following solutons:
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document.write( "  For these solutions to exist, the discriminant  should not be a negative number.
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document.write( "  First, we need to compute the discriminant : .
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document.write( "  Discriminant d=1936 is greater than zero. That means that there are two solutions: .
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document.write( "    Quadratic expression  can be factored:
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document.write( "  Again, the answer is: 7, -7.66666666666667.\n" );
document.write( "Here's your graph:
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