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put this solution on YOUR website!Show that b^3+b+1=a^3 has no positive integer solutions.
LET US TEST IT UNDER 2 STEPS...
1..POSITIVE ROOT EXISTS...AND THEN
2..IT SHOULD BE AN INTEGER.
F(B)= b^3+b+1-a^3=0..
USING THE RULE OF SIGNS ,FOR THIS TO HAVE A POSITIVE ROOT 1-A^3 HAS TO BE NEGATIVE,SINCE B IS POSITIVE.
THAT IS A>1..LET US ASSUME SO..THEN WE SHALL HAVE
BASIS .....A>1...
F(B)= b^3+b-(A^3-1)=0...WHERE A^3-1 IS POSITIVE.
NOW FOR THE ROOT TO BE RATIONAL ,THE ROOT HAS TO BE FACTOR OF A^3-1...SINCE COEFFICIENT OF B IS 1,WE SHALL HAVE AN INTEGER ROOT IF WE GET A RATIONAL ROOT.
FACTORS OF A^3-1 ARE (A-1) AND A^2+A+1...WHICH ARE NO MORE REDUCIBLE.TESTING THEM
I..FIRSTLY FOR A-1..WE GET...
F(A-1)=(A-1)^3+A-1-A^3+1=A^3-1-3A^2+3A+A-1-A^3+1=-3A^2+4A-1 SHOULD EQUAL ZERO
=-3A^2+3A+A-1=3A(1-A)-1(1-A)=0
(3A-1)(1-A)=0
CASE 1..............3A-1=0...OR....A=1/3 IS NOT POSSIBLE AS A>1 AS PER OUR BASIS FOR THE ROOT TO BE POSITIVE
CASE 2....1-A=0.....OR...A=1....IS NOT POSSIBLE AS A>1 AS PER OUR BASIS FOR THE ROOT TO BE POSITIVE.
SO A-1 IS NOT A ROOT..
II.....SECONDLY FOR A^2+A+1...WE GET...
F(A^2+A+1)=(A^2+A+1)^3+A^2+A+1-A^3+1 SHOULD EQUAL ZERO OR A VALUE OF A>1
WE FIND THAT THERE IS ONLY ONE NEGATIVE NUMBER -A^3,WHICH GETS CANCELLED OUT BY +A^3 IN THE EXPANSION OF (A^+A+1)^3....LEAVING ALL OTHERS AS SUM OF POSITIVE TERMS WHICH CAN NEVER EQUAL ZERO.HENCE A^2+A+1 IS ALSO NOT A FACTOR.
HENCE F(B) HAS NO RATIONAL ROOT FOR A>1
HENCE F(B) HAS NO POSITIVE INTEGRAL ROOT
