How do you factor (t-1)^6 + (1-t)^3?

Question

Hey, i need help factoring this : (t-1)^6 + (1-t)^3
Please SHOW work and EXPLAIN. Thanks in advance.

Answer 1

On thing to notice is that (t-1) = -(1-t)

Begin by rewriting the first:

[-1* (1-t)]^6 + (1-t)^3

Next (a*b)^c is the same as a^c + b^c:

(-1)^6 * (1-t)^6 + (1-t)^3

And (-1)^6 is 1, so you have simplified it to:
(1-t)^6 + (1-t)^3.

Now factor out (1-t)^3:
(1-t)^3[(1-t)^3 + 1]

Answer 2

Remember that:
t -1 = -1 * (1 - t)

Therefore substitute this into the second term.

(t-1)^6 + [-1 * (t-1)]^3
(t-1)^6 + (-1)^3 * (t-1)^3
(t-1)^6 - (t-1)^3 ---> now factor out the (t-1)^3

(t-1)^3 * [ (t-1)^3 - 1 ]

Answer 3

(t-1)^6 + (1-t)^3
= (t-1)^6 - (t-1)^3
= [(t-1)^3][(t-1)^3-1]
= [(t-1)^3](t-2)[(t-1)^2+(t-1)+1]
= [(t-1)^3](t-2)(t^2-t+1)

Answer 4

One possibility is to use the sum of cubes formula, but I think a better way might be to say:

(1-t) = -(t-1) so
(t-1)^6 + (1-t)^3 = (t-1)^6 - (t-1)^3 = (t-1)^3[(t-1)^3 - 1]

Answer 5

(t-1)^6 + (1-t)^3 =
(t-1)^6 - (t-1)^3 =
{(t-1)^3}^2 - (t-1)^3 =
(t-1)^3{(t-1)^3 - 1} =
(t-1)^3{(t-1)^3 - 1^3} =
(t-1)^3{[(t-1) - 1][t^2 + t + 1]}
(t-1)^4 x ( t^2 + t + 1)

Th