# Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 39

$2y(3y-4)(9y^2+12y+16)$

#### Work Step by Step

Factoring the $GCF= 2y ,$ the given expression, $54y^4-128y ,$ is equivalent to \begin{array}{l} 2y(27y^3-64) .\end{array} Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the expression, $8(8x^6-t^6) ,$ is \begin{array}{l} 2y(3y-4)[ (3y)^2-(3y)(-4)+(-4)^2] \\\\= 2y(3y-4)(9y^2+12y+16) .\end{array}

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