If I have (m^6 - n^6) -
I know I can factor it as a difference of two squares:
(m^3 + m^3)(m^3 - m^3)
and then as the sum AND difference of two cubes:
(m+n)(m^2 - mn + n^2) (m-n) (m^2 + mn + n^2)
BUT what if I chose to factor it as a difference of 2 cubes FIRST and I get:
(m^2 - n^2)[(m^2)^2 + [(m^2)(n^2)] + (n^2)^2]
Why can't I figure out how to factor the trinomial part and still get the same ENDING answer as the first method? I can see that the BINOMIAL is a difference of 2 squares and see that portion in the first answer - however the trinomial is NOT a perfect square (and I do not see any error I have made?)
I know the first method is EASIER and the method of choice but shouldn't I be able to do it the second way too??? And still get the same answer??? in the same simplified form???
Please HELP!!!!!!!!!!!
I know I can factor it as a difference of two squares:
(m^3 + m^3)(m^3 - m^3)
and then as the sum AND difference of two cubes:
(m+n)(m^2 - mn + n^2) (m-n) (m^2 + mn + n^2)
BUT what if I chose to factor it as a difference of 2 cubes FIRST and I get:
(m^2 - n^2)[(m^2)^2 + [(m^2)(n^2)] + (n^2)^2]
Why can't I figure out how to factor the trinomial part and still get the same ENDING answer as the first method? I can see that the BINOMIAL is a difference of 2 squares and see that portion in the first answer - however the trinomial is NOT a perfect square (and I do not see any error I have made?)
I know the first method is EASIER and the method of choice but shouldn't I be able to do it the second way too??? And still get the same answer??? in the same simplified form???
Please HELP!!!!!!!!!!!