## Algebra and Trigonometry 10th Edition

$(x-\frac{3}{4})(x^2+\frac{3}{4}x+\frac{9}{16})$
$x^3-\frac{27}{64}=x^3-(\frac{3}{4})^3=(x-\frac{3}{4})[x^2+x(\frac{3}{4})+(\frac{3}{4})^2]=(x-\frac{3}{4})(x^2+\frac{3}{4}x+\frac{9}{16})$
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