Question

can any one tell me , why is this so

Answer 1

might wanna factor that a^3 - b^3 to compare the sides with

Answer 2

i actually didn't understand the question , only i understand is that a^3-b^3 formula is not equal to what is written on RHS

Answer 3

\[a^3-b^3 = (a-b)(a^2+ab+b^2)\]

\[(a-b)(a^2+ab+b^2) = 1+3ab\]
\[(a-b)(a^2+ab+b^2) = (a-b)(\frac{1}{a-b}+\frac{3ab}{a-b})\]
\[a^2+ab+b^2+ab = \frac{1}{a-b}+\frac{3ab}{a-b}+ab\]
\[a^2+2ab+b^2 = \frac{1+(3a-3b)ab}{a-b}\]
not too sure where this may lead

Answer 4

the left side is a perfect square, so the right side has to be greter or equal to 0

Answer 5

okay

Answer 6

\[\frac{1+(3a-3b)ab}{a-b}\ge0~:~if~a>b,~then~(a-b)~is~+\]
\[1+(3a-3b)ab\ge0\]

\[\frac{1+(3a-3b)ab}{a-b}\ge0~:~if~a<b,~then~(a-b)~is~-\]
\[1+(3a-3b)ab\le0\]

Answer 7

\[1+3(a-b)ab\ge0~:~a>b\]\[(+)ab\ge-\frac{1}{3}\]

\[1+3(a-b)ab\le0~:~a<b\]\[(-)ab\le-\frac{1}{3}\]\[ab\ge\frac{1}{3}\]

might have stayed on track with that

Answer 8

@amistre64
thank u
i understand a lit bit about the question,

Answer 9

i hope it helps becasue thats all i think i can get out of it, so long as a read it correctly to begin ewith

Answer 10

as far as I can understand it, it's asking if "a" and "b" have be 1 number apart, meaning say, a=5 then b=a-1=4, or can they have more "difference", like say a=7 then b=25, instead of b=a-1=6

Answer 11

well, if a is 7 b can't be 25 :|, but something less like say 2 or 3, not 6 :)

Answer 12

let \(a-b=m\) u will get\[3mb^2-3mb+m^3-1=0\]a quadratic in terms of \(b\) so in order for \(b\) to be an integer number the determinant of the quadratic must be greater than or equal to zero, we must have\[-12m^4+9m^2+12m\ge0\]its possible only if \(m=1\)