Question

If $(0.2{)}^3-(0.3{)}^3+(0.1{)}^3=-m$. Find $1000m$. Without actually calculating cubes,

Answer 1

$(0.2{)}^3-(0.3{)}^3+(0.1{)}^3$

$a^3-b^3=(a-b)(a^2+b^2+ab)$

$\left[\right(0.2{)}^3-\left(0.3{)}^3\right]+(0.1{)}^3$

$\Rightarrow \left[\right(0.2-0.3\left)\right\{(0.2{)}^2+(0.3{)}^2+\left(0.2\right)\left(0.3\right)\left\}\right]+1(0.1{)}^3$

$\Rightarrow [-(0.1\left)\right(0.04+0.09+0.06\left)\right]+0.001$

$\Rightarrow [-0.019+0.001]$

$\Rightarrow -0.018$

$\therefore m=0.018$

$1000m=18$