Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 44

$(p^2-q^2)(p^4+p^2q^2+q^4)$

Work Step by Step

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $p^6-q^6 ,$ is \begin{array}{l} (p^2-q^2)[ (p^2)^2-(p^2)(-q^2)+(-q^2)^2] \\\\= (p^2-q^2)(p^4+p^2q^2+q^4) .\end{array}

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