Question

factorise 1/27 - x³/64 ​

Answer 1

Answer:

$(\frac{1}{3}-\frac{x}{4})(\frac{1}{9} + \frac{x}{12} + \frac{x^2}{16})$

Step-by-step explanation:

Given:

$\frac{1}{27} - \frac{x^3}{64} \\\\=(\frac{1}{3})^3 - (\frac{x}{4})^3\\\\=a^3 -b^3= (a-b)(a^2 +ab+b^2)\\\\=(\frac{1}{3}-\frac{x}{4})([\frac{1}{3}]^2 + [\frac{1}{3}][\frac{x}{4}] + [\frac{x}{4}]^2)\\\\=(\frac{1}{3}-\frac{x}{4})(\frac{1}{9} + \frac{x}{12} + \frac{x^2}{16})$