Last 3 digits

$\frac{877^3-123^3}{877-123} = \frac{(877-123)(877^2+123^2+877\times123)}{877-123}$

$877^2+123^2+2\times877\times123-873\times123 = (877+123)^2-877.123 = 1000^2 - 877.123$

Since we need to find last three digits our working modulo will be 1000

$1000^2-877\times123 \equiv 1000-871 \equiv 129 \pmod{1000}$

(877³-123³)/(877-123) = [(877-123)(877²+877.123+123²)]/(877-123) = (877²+877.123+123²) = (877+123)²-877.123 =1000000 -107871=892129

Using the identity that :

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

We can further transform this for our convenience as :

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

Now we can write the equation as :

$\begin{aligned} \dfrac{877^3 - 123^3}{877 - 123} &= \dfrac{( 877 - 123 ) [ ( 877 + 123 )^2 - 877 \times 123 ]}{(877 - 123)} \\ &= (1000)^2 - 877 \times 123 \\ &= (1000)^2 - (1000 - 877) \times 877 \\ &= (1000)^2 - 2(1000)(877) + (877)^2 + (1000)(877) \\ &= (1000-877)^2 + 877000 \\ &= (123)^2 + 877000 \\ &= 15129 + 877000 \end{aligned}$

Thus the last 3 digits will be = $\boxed{129}$

Cheers!

Use calculator. problem officer?