Question Number 183363 by Shrinava last updated on 25/Dec/22
$$\mathrm{Find}: \\ $$$$\mathrm{2003}\centerdot\mathrm{2005}^{\mathrm{3}} −\mathrm{2004}\centerdot\mathrm{2002}^{\mathrm{3}} \\ $$
Answered by Frix last updated on 25/Dec/22
$$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} −\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{8}{x}^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{8}×\frac{\mathrm{4007}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }=\mathrm{4007}^{\mathrm{3}} \\ $$