Can You Make This Complex Calculation Simple?

First, we see that $8=2^3$ .By factoring $500^3-2^3$ , we get $(500-2)(500^2+2*500+2^2)$ . We notice that $(500^2+2*500+2^2)$ is the same as the denominator, so we can divide that out leaving us just with $500-2=\boxed{498}$

Let $a=500$ , $b=2$ . Then, we must find $\dfrac{a^3-b^3}{a^2+ab+b^2}$ However, we know the difference of cubes factorization , which says that $(a-b)(a^2+ab+b^2)=a^3-b^3$ Therefore, the answer is $a-b=500-2=\boxed{498}$ .

Note that

$500^{3}-8$

$= 500^{3}-2^{3}$

$= (500-2)(500^{2}+2 \times 500 + 2^{2})$

$= (500-2)(500^{2} + 2 \times 500 + 4)$

Hence,

$\dfrac{500^{3}-8}{500^{2}+2 \times 500 + 4}$

$= \dfrac{(500-2)(500^{2}+2 \times 500 + 4)}{500^{2}+2 \times 500 + 4}$

$= 500 -2$

$= \boxed{498}$

Calculator.

(500^3-2^3)/(500^2+2 500+2^2)=(500-2)(500^2+2 500+2^2)/ (500^2+2*500+2^2)=500-2=498 here we use the identity (a^3-b^3)=(a-b)(a^2+ab+b^2)

500^3 - 8=124999992 500^2+2*500+4=251004 124999992/251004=498

(500^3-8)/500^2+2 500+4), here, (500^3-8) is to be (500^3-2^3) like this so expansion of a^3-b^3 is (a-b)(a^2+ab+b^2) in this way this sum write like this (500-2)(500^2+2 500+4)/(500^2+2*500+4) so cancel remaining (500-2) result is 498

answer for this problem:

Nu:500cube-8=(500 500 500)-8=125000000-8=124999992

De:500square+2 100+4=(500 500)+1000+4=250000+1004=251004

Nu/De=124999992-251004=498