document.write( "Question 390658: find 3 consecutive integers such that the product of all 3, decreased by the cube of the first, is 33 \n" ); document.write( "

Algebra.Com's Answer #277078 by checkley79(3341) $\"\"$ $\"About$

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Let x, x+1, x+2 be the 3 integers.
\n" ); document.write( "x(x+1)(x+2)-x^3=33
\n" ); document.write( "(x^2+x)(x+2)-x^3=33
\n" ); document.write( "x^3+x^2+2x^2+2x-3x^3=33
\n" ); document.write( "3x^2+2x=33
\n" ); document.write( "3x^2+2x-33=0
\n" ); document.write( "(3x+11)(x-3)=0
\n" ); document.write( "x-3=0
\n" ); document.write( "x=3 ans.
\n" ); document.write( "Proof:
\n" ); document.write( "3(3+1)(3+2)-3^3=33
\n" ); document.write( "3*4*5-27=33
\n" ); document.write( "60-27=33
\n" ); document.write( "33=33 \n" ); document.write( "

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