Precalculus (6th Edition)

RECALL: (i) $a^3-b^3 = (a-b)(a^2+ab+b^2)$ (ii) $a^3+b^3=(a+b)(a^2-ab+b^2)$ (a) $8x^3-27=(2x)^2-3^2$ Thus, using the formula in (1) above gives: $8x^3-27 \\=(2x)^3-3^3 \\=(2x-3)[(2x)^2+(2x)(3)+3^2] \\=(2x-3)(4x^2+6x+9)$ Thus, the answer is B. (b) $8x^3+27=(2x)^2+3^2$ Thus, using the formula in (2) above gives: $8x^3+27 \\=(2x)^3+3^3 \\=(2x+3)[(2x)^2-(2x)(3)+3^2] \\=(2x+3)(4x^2-6x+9)$ Thus, the answer is C. (c) $27 - 8x^3=3^3-(2x)^3$ Thus, using the formula in (1) above gives: $27-8x^3 \\=3^3-(2x)^3 \\=(3-2x)[3^2+(3)(2x)+(2x)^2] \\=(3-2x)(9+6x+4x^2)$ Thus, the answer is A.