2006-10-17 11:47:11 · 5 answers · asked by Tony K 1 in Science & Mathematics ➔ Mathematics
(x-1)(x-1)(x-1) is the long way of saying it...multiply and you'll get your answer.
2006-10-17 11:48:37 · answer #1 · answered by Shaun 4 · 0⤊ 0⤋
(X-1)^3 = (X-1)^2 . (X-1)
= (X-1)(X^2 + 1 - 2X)
= X^3 + X - 2X^2 - X^2 - 1 + 2X
=X^3 + 3X - 3X^2 - 1
=(X^3 - 1) + 3X (X-1)
2006-10-17 18:54:14 · answer #2 · answered by shamit_shailesh 1 · 0⤊ 0⤋
(x-1) (x-1) (x-1)
(x-1) (x^2 - 2x +1)
x^3 - 2x^2 + x - x^2 +2x -1
x^3 - 3x^2 + 3x - 1
2006-10-17 18:53:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(x-1)^3 = (x-1)(x-1)(x-1)
Use foil or the binomial theorem.
2006-10-17 18:49:52 · answer #4 · answered by MsMath 7 · 0⤊ 0⤋
you better know the Newton triangle:
(a+b)^0=1
(a+b)^1=a+b
(a+b)^2=a^2+2ab+b^2
(a+b)^3=a^3+3a^2*b+3a*b^2+b^3
(a+b)^4=a^4+4a^3*b+6a^2*b^2+4a*b^3+b^4
(a+b)^5=a^5+5a^4*b+10a^3*b^2+10a^2*b^3+5a*b^4+b^5
...
(a+b)^n=a^n+na^(n-1)*b+(nC2)a^(n-2)*b^2+(nC3)a^(n-3)*b^3+...+[nC(n-1)]a*b^(n-1)+b^n
where nCr=n!/[(n-r)!*r!]
2006-10-17 19:39:04 · answer #5 · answered by Anonymous · 0⤊ 0⤋