document.write( "Question 783031: If the product of the integers a, b, and c is 1, then what is the smallest possible value of (a+1)(b+1)(c+1)?
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Algebra.Com's Answer #496513 by tommyt3rd(5050)\"\" \"About 
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Expanding and simplifying...
\n" ); document.write( "(a+1)(b+1)(c+1)=
\n" ); document.write( "(a+1)(bc+b+c+1)=
\n" ); document.write( "abc+ab+ac+bc+a+b+c+1=
\n" ); document.write( "ab+bc+ac+a+b+c+2\r
\n" ); document.write( "\n" ); document.write( "Since their product is 1, the only possibilities are a,b,c={-1,1} (why?)
\n" ); document.write( "Furthermore, there must be an odd number of positives. To find the minimum we can ignore the case a=b=c=1 - that would be the maximum.\r
\n" ); document.write( "\n" ); document.write( "Case1
\n" ); document.write( "if say a=b=1, c=-1, then\r
\n" ); document.write( "\n" ); document.write( "ab+bc+ac+a+b+c+2=
\n" ); document.write( "1-1-1+1+1-1+2=
\n" ); document.write( "2\r
\n" ); document.write( "\n" ); document.write( "Case2
\n" ); document.write( "if say c=1, a=b=-1, then\r
\n" ); document.write( "\n" ); document.write( "ab+bc+ac+a+b+c+2=
\n" ); document.write( "1-1-1-1-1+1+2=
\n" ); document.write( "0\r
\n" ); document.write( "\n" ); document.write( "Minimum value is 0
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