Question

Could anybody help finding the two factors of this equation: p(x)=1+(x-2)^3
one of the factor is (x-1)

Thanks!

Answer 1

1 = 1^3

you can write the right side into the form a^3 + b^3

a = 1
b = x-2

then use the sum of cubes factoring formula

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

Answer 2

Actually I think ^3 is just for (x-2) but not 1

Answer 3

what do you mean?

Answer 4

p(x)=1+(x-2)^3 is the same as p(x)=1^3+(x-2)^3

since 1 = 1^3

Answer 5

1^3 = 1*1*1 = 1

Answer 6

oh true!

Answer 7

omg I got it!!!! So smart!
Thank you very much!

Answer 8

you're welcome