Subtracting integers with powers

AI Thread Summary
To evaluate 6667² - 3333² without a calculator, the difference of two squares formula, a² - b² = (a + b)(a - b), can be applied. Using this formula, the expression simplifies to (6667 + 3333)(6667 - 3333), which equals 10000 * 3334. This method is more efficient than subtracting the integers directly and adding zeros. Understanding this concept is crucial for solving similar problems effectively.
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Homework Statement


Hey, sorry about this. Its really obvious, i think there's just some really simple way to do it. Its annoying me, my teacher couldn't do it either!

Evaluate 6667²-3333² (without a calculator)

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The Attempt at a Solution



I know there is just some really obvious way to do it. It was only worth 2 marks. Checking the answer on a calculator it was 33340000 and 6667-3333=3334 so i wasnt sure if that was how you did it, found the answer to the integers minus each other and add zeros. But i tried other examples and it didnt hold true.

Thanks
 
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At some point in class, you should have covered what the difference of two squares is. Just to remind you, it looks something like this:
a^2-b^2=(a+b)*(a-b)

I hope that helps.
 
You in general can also try to factorize the terms and calculate and simplify that way; but in the example you gave, knowing the difference of two squares is more efficient as in post #2.
 
Awesome, cheers guys!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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