Question

Without actually calculating the cubes, find,

(1/2)^3 + (1/3)^3 - (5/6)^3

Answer 1

$( \frac{1}{2})^{3} + (\frac{1}{3})^3 - (\frac{5}{6})^3 \\ \\ we \ use\ a^3+b^3 = (a+b)(a^2-ab+b^2), \ \ a^3-b^3 = (a-b)(a^2+ab+b^2) \\ \\ (\frac{1}{2}+\frac{1}{3})(\frac{1}{2^2}-\frac{1}{2*3}+\frac{1}{3^2}) - (\frac{5}{6})^3 \\ \\ \frac{5}{6} (\frac{9-6+4}{4*9}) - (\frac{5}{6})^3 \\ \\ \frac{5}{6} (\frac{7}{6^2}) - (\frac{5}{6})^3 \\ \\ \frac{5}{6} [\frac{7}{6^2} - \frac{5^2}{6^2} ] \\ \\ \frac{5}{6} [\frac{7-25}{6^2}] \\ \\ \frac{- 5 * 18}{6*6*6} \\ \\ - \frac{5}{12}$

Answer 2

Use the identity a³ + b³ + c³ = 3abc if a + b + c = 0

Here a = (1/2), b = (1/3), c = (-5/6)
a + b + c = (1/2) + (1/3) + (-5/6) = 0
Therefore (1/2)^3 + (1/3)^3 - (5/6)^3 = 3 x (1/2)x(1/3)x(-5/6) = - 5/12