How do you factor ${(t + u)}^{3} - 64$ $(t+u)^3-64$ ?

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Answer 1 · 2015-05-07T15:28:13

Use a³ - b³ = (a - b)(a² + ab + b²)

$a -] {(t + u)}^{3}$ $a-](t+u)^3$
$b -] 64$ $b-]64$

$therefore (t+u)^3-64$
$=(t+u−64){(t+u)^2+(t+u)(64)+64^2}$

How do you factor (t+u)^3-64(t+u)3−64(t+u)^3-64?