## College Algebra 7th Edition

Published by Brooks Cole

# Chapter P, Prerequisites - Chapter P Review - Exercises - Page 77: 63

#### Answer

$(x-3)(x^2+3x+9)$

#### Work Step by Step

Since $27=3^3$, the given expression is equivalent to: $=x^3-3^3$ The expression above is a difference of two cubes. Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x$ and $b=3$ to obtain: $=(x-3)(x^2+x(3) + 3^2) \\=(x-3)(x^2+3x+9)$

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