document.write( "Question 353112: A rectangular box has sides with lengths x, x+1 and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box. \n" ); document.write( "
Algebra.Com's Answer #252335 by Jk22(389)\"\" \"About 
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Methods
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\n" ); document.write( "a) we decompose 504 in factor : 504/2=252, 252/2=126, 126/2=63, 63=7*9
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\n" ); document.write( "hence 504=2*2*2*7*9=8*7*9 which is the result
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\n" ); document.write( "b) Solving the equation :
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\n" ); document.write( "let y=x+1, we get : (y-1)y(y+1)=504
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\n" ); document.write( "(y^2-1)y=504 => y^3-y-504=0
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\n" ); document.write( "let y=p+q, (p+q)^3-(p+q)-504=0
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\n" ); document.write( "p^3+3p^2q+3pq^2+q^3-(p+q)-504=0
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\n" ); document.write( "p^3+q^3+3pq(p+q)-(p+q)-504=0
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\n" ); document.write( "p^3+q^3+(p+q)(3pq-1)-504=0
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\n" ); document.write( "let p^3+q^3=504
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\n" ); document.write( "3pq=1 => p=1/3q => 1/(27q^3)+q^3=504
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\n" ); document.write( "q^6-504q^3+1/27=0, q^3=t => t^2-504t+1/27=0
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\n" ); document.write( "t=(504 +/- sqrt(504^2-4/27))/2
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\n" ); document.write( "q=t^(1/3) and y=p+q=1/3/t^(1/3)+t^(1/3)=8=y=x+1, hence 504=7*8*9 \n" ); document.write( "
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