Question 371134
We can factor this two ways.
1. Factor out a -1:
{{{(-1)(c^3+125)}}}
then factor as a sum of cubes (since {{{125 = 5^3}}}). The pattern for sum of cubes is: {{{a^3+b^3 = (a+b)(a^2-ab+b^2)}}}. Using this on {{{c^3+125}}} we get:
{{{(-1)((c)+(5))((c)^2 - (c)(5) + (5)^2)}}}
which simplifies to:
{{{(-1)(c+5)(c^2 - 5c + 25)}}}<br>
2. Factor as a difference of cubes (since {{{-c^3 = (-c)^3}}} and 125 is still a perfect cube. The pattern for difference of cubes is: {{{a^3-b^3 = (a-b)(a^2+ab+b^2)}}}. Using this on {{{-c^3-125}}} we get:
{{{((-c)-(5))((-c)^2+(-c)(5)+(5)^2)}}}
which simplifies to:
{{{(-c-5)(c^2-5c+25)}}}<br>
Either answer is correct. They are equivalent.