I believe its in relation to pascals triangle, but this has thrown me.
2007-03-18 01:15:15 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1 2 1
1 3 3 1
1 4 6 4 1
(x-1/x)^4 = x^4 - 4 x^3/x + 6 x^2/x^2 - 4x/x^3 + 1/x^4
= x^4 -4x^2 +6 -4/x^2 +1/x^4
2007-03-18 01:19:06 · answer #1 · answered by hustolemyname 6 · 0⤊ 0⤋
By the binomial theorem
(x-1/x)^4 = x^4 - 4x^3/x + 6x^2/x^2 - 4x/x^3 + 1/x^4 = x^4 - 4x^2 + 6 - 4/x^2 + 1/x^4
2007-03-18 08:27:12 · answer #2 · answered by physicist 4 · 0⤊ 0⤋
(x - 1/x)^4 = ((X^2-1)/X)^4
Using Pascal, we get:
(x^8-4x^6+6x^4-4x^2+1)/x^4 or
x^4 - 4x^2 + 6 - 4/x^2 + 1/x^4
2007-03-18 08:24:46 · answer #3 · answered by blighmaster 3 · 0⤊ 0⤋
(x-1/x)^4
=1/x^4{x^2-1)^4
=1/x^4
{x^8-4x^6+6x^4-4x^2+1)
=x^4-4x^2+6-4/x^2+1/x^4
i hope that this helps
2007-03-18 09:50:22 · answer #4 · answered by Anonymous · 0⤊ 0⤋