To find the value of $99^3$, we can use the algebraic identity for the cube of a binomial difference: $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.
We rewrite $99$ as $(100 - 1)$. In this case, $a = 100$ and $b = 1$.
Sum the results from the previous step:
$99^3 = 1,000,000 - 30,000 + 300 - 1$
$99^3 = 970,000 + 300 - 1$
$99^3 = 970,300 - 1$
$99^3 = 970,299$
The value of $99^3$ is $970,299$.
Evaluate \(21^3 + (-2)^3 + (-19)^3\)
Two different positions of the same dice are shown, the six face of which are numbered from 1 to 6. Select the number that will be on the face opposite to the face having the number '3'.
