Question

If $\alpha =(0.2{)}^3-(0.3{)}^3+(0.1{)}^3$, without actually calculating the cubes, find the value of $-1000\alpha$ :

Answer 1

We know that,
$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
If $a+b+c=0$, then $a^3+b^3+c^3=3abc$
Now, given $(0.2{)}^3-(0.3{)}^3+(0.1{)}^3$
Here, $a=0.2$, $b=-0.3$, $c=0.1$ and $a+b+c=0.2+(-0.3)+0.1=0$
Therefore
$\alpha =(0.2{)}^3-(0.3{)}^3+(0.1{)}^3=3(0.2\left)\right(-0.3\left)\right(0.1)$
$=-0.018$

$\therefore -1000\alpha =18$