<s>[INST] Solve for $x$ where $x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$ Consider the following equality:
$$x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$$
Solve for $x$.
The only thing I noticed is the first part like $(1-2)$,$(3-4)$ gives us $-1$ but then I just don't see what the trick behind this problem is.
 [/INST] Hint: have you ever tried to look at a formula like this: $$ {a^3 - b^3 \over a-b} = (?)(?)$$ ? Just expand and observe the two occuring factors and look like the astronomers over long distances ...              
[update] 
As $ {a^3-b^3 \over a - b }= a^2+ab+b^2$ and thus $$  (a-b)(a^2+ab+b^2) = a^3-b^3 $$the problem-formula is equivalent $(a^3-b^3) + (b^3-c^3) + ... +(x^3-y^3) +  (y^3-z^3) = a^3-z^3 $ and comes out to be $$x = 1^3 - 50^3 = -124999 $$ 
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