it appears to be a very easy question......:) here is the probable solution....:
log[(a+b)/3]=(log a+log b)/3
=>log[(a+b)/3]=log(ab)/3
=>3log[(a+b)/3]=log(ab)
=>log[{(a+b)/3}^3]=log(ab)...........taking antilog,.......
=>[(a+b)^3]/[27]=ab
=>a^3+b^3+3ab(a+b)=27ab
=>so the integer solutions are [a=2 & b=4] ; [a=4 &b=-16] ; [a=4 & b=2].....BUT YOU SEE "b" CAN NOT TAKE NEGATIVE VALUE INSIDE LOGARITHM.....SO THE SOLUTION SHOULD BE.....[a=2 & b=4] AND [a=4 & b=2]..........................
HOPE THIS HELPS.........:))

ha ha....!! ...i thought if anyone asks me that step by step solution.....i would crumble....!!!.....then to find the solution of a^3+b^3+3ab(a+b)=27ab i only put my graphing knowledge.........BUT EASY TO SAY YOU MAY FIND THE SOLUTION WITHOUT APPLYING YOUR BRAIN EVEN BY CLICKING HERE SIMPLY.......i only used this website to check my answer whether it is right or wrong......:-)

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TopNewestHow do we get the integer solutions ?????

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it appears to be a very easy question......:) here is the probable solution....: log[(a+b)/3]=(log a+log b)/3

=>log[(a+b)/3]=log(a

b)/3log[(a+b)/3]=log(a=>3

b)b)...........taking antilog,....... =>[(a+b)^3]/[27]=a=>log[{(a+b)/3}^3]=log(a

bab=>a^3+b^3+3

(a+b)=27ab=>so the integer solutions are [a=2 & b=4] ; [a=4 &b=-16] ; [a=4 & b=2].....BUT YOU SEE "b" CAN NOT TAKE NEGATIVE VALUE INSIDE LOGARITHM.....SO THE SOLUTION SHOULD BE.....[a=2 & b=4] AND [a=4 & b=2].......................... HOPE THIS HELPS.........:))

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how did you directly deduce the integer solns from the step:=>a^3+b^3+3ab(a+b)=27ab

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ha ha....!! ...i thought if anyone asks me that step by step solution.....i would crumble....!!!.....then to find the solution of a^3+b^3+3ab(a+b)=27ab i only put my graphing knowledge.........BUT EASY TO SAY YOU MAY FIND THE SOLUTION WITHOUT APPLYING YOUR BRAIN EVEN BY CLICKING HERE SIMPLY.......i only used this website to check my answer whether it is right or wrong......:-)

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How do we get the integer solutions ?????

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