1 = 1.25 ^ (2) - 0.75 ^ (2)
similarly express 1 as difference of two cubes.
2006-07-21 18:55:47 · 6 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
Can't be done with rational numbers, because if:
1 = (p/q)^3 - (r/s)^3,
then (sq)^3 = (ps)^3 - (rq)^3
and we'd have a non-trivial solution to the equation x^3 + y^3 = z^3.
By Fermat's last theorem for n=3, there are no such solutions.
So there are no (non-trivial) solutions to your equation.
[ As one poster noted, 1 = 1^3 - 0^3. Also, 1 = 0^3 - (-1)^3.
Those are 'trivial' solutions. ]
2006-07-21 18:59:38 · answer #1 · answered by thomasoa 5 · 1⤊ 0⤋
The expression is true.
1= 1.25^(2) - 0.75^(2)
1= 1.56250 - 0.56250
1 =1
This is a difference of two squares, exponents being 2.
A cube's exponent = 3.
2006-07-22 02:03:42 · answer #2 · answered by just me 4 · 0⤊ 0⤋
1^3 -0^3, what's wrong with that?
there are infitely many solutions, for any x larger than 1 you can find a second that would give one as difference of cubes
( y = cube-root(1-x^3) )
or you want numbers symmetric around 1?
solve: (1+x)^3 - (1-x)^3 = 1
1 +3x +3x^2 +x^3 - 1+3x -3x^2 +x^3 = 1
2x^3 +6x -1 = 0
this is has a root between 0 and 1, have fun finding it - I got better stuff to do
2006-07-22 01:58:57 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1=1(I want to learn)
2006-07-22 02:00:58 · answer #4 · answered by Rim 6 · 0⤊ 0⤋
yeah ok, sorry math was not my subject.
2006-07-22 01:59:11 · answer #5 · answered by donna 4 · 0⤊ 0⤋
I'm pretty sure that's not possible...
2006-07-22 02:01:43 · answer #6 · answered by romantemple16 2 · 0⤊ 0⤋