How do you factor $27 - t^{3}$ $27-t^3$ ?

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Answer 1 · 2015-10-09T05:43:35

$(3 - t) (9 + 3 t + t^{2})$ $(3-t)(9+3t+t^2)$

$27 - t^{3} \Rightarrow$ $27-t^3=>$ use difference of cubes formula:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \Rightarrow$ $a^3-b^3=(a-b)(a^2+ab+b^2)=>$ so in this case we have:

$(3^{3} - t^{3}) = (3 - t) (9 + 3 t + t^{2})$ $(3^3-t^3)=(3-t)(9+3t+t^2)$

How do you factor 27-t^327−t327-t^3?