If you add and subtract x^2 from the numerator you can split up the fraction in to 1 + (x^5 - x^2) / (x^2+x+1) = 1 + x^2 * (x^3 - 1) / (x^2+x+1) = 1 + x^2 * (x - 1)(x^2 + x + 1)/ (x^2+x+1) = 1+ x^3 - x^2 Final result x^3 - x^2 + 1 There's also a way of doing this by factoring the numerator but it's gonna be a bit of brute force and systems and a whole lot of work, and maybe some guessing. It's a nice problem that rewards you for good observation making.
I just did long division. looks something like this. the x^2 + x + 1 goes into it x^3 number of times leaving remainder of -x^4 - x^3 + x + 1 which then goes into it -x^2 number of times leaving remainder x^2 + x + 1 which goes into it 1 time. Final answer is x^3 - x^2 + 1
Multiply both top and bottom by (x-1). The top is x^6-1 and the bottom is x^3-1. Factori both into the complex roots of one and watch the carnage of cancellation.
The numerator will have x^6-1 and two extra terms, but this approach still works well. Num = x^6 - 1 - x^5 + x^2 = (x^3+1)(x^3-1) - x^2 (x^3-1) = (x^3+1-x^2)(x^3-1)
If you add and subtract x^2 from the numerator you can split up the fraction in to
1 + (x^5 - x^2) / (x^2+x+1) =
1 + x^2 * (x^3 - 1) / (x^2+x+1) =
1 + x^2 * (x - 1)(x^2 + x + 1)/ (x^2+x+1) =
1+ x^3 - x^2
Final result x^3 - x^2 + 1
There's also a way of doing this by factoring the numerator but it's gonna be a bit of brute force and systems and a whole lot of work, and maybe some guessing. It's a nice problem that rewards you for good observation making.
I just did long division. looks something like this. the x^2 + x + 1 goes into it x^3 number of times leaving remainder of -x^4 - x^3 + x + 1 which then goes into it -x^2 number of times leaving remainder x^2 + x + 1 which goes into it 1 time.
Final answer is x^3 - x^2 + 1
My method too, appeared quite simple
Just add (and subtract) x^4+x^3+x^2...
Yeah either that or synthetic division according to your mood
Add and subtract x^2 in numerator to get x^2(x-1) +1
Multiply both top and bottom by (x-1). The top is x^6-1 and the bottom is x^3-1. Factori both into the complex roots of one and watch the carnage of cancellation.
The numerator will not be X^6-1
@@yuryp6975 Well I feel stupid. Thanks--- I was thinking x^5+x^4+x^3+x^2+x+1 just missed a few terms.
@@yuryp6975 Your upper case letters represent different variables from the user's lower case letters.
The numerator will have x^6-1 and two extra terms, but this approach still works well.
Num = x^6 - 1 - x^5 + x^2
= (x^3+1)(x^3-1) - x^2 (x^3-1)
= (x^3+1-x^2)(x^3-1)
x^3-x^2+1
Wow
*(x⁵ +x +1) / (x² +x +1) = (x⁵ +x⁴ +x³ +x² +x +1 - x⁴ - x³ - x²) / (x² +x +1) = [x³(x² +x +1) +1(x² +x +1) - x²(x² +x +1)] / (x² +x +1) = **-(x² +x +1)-** • (x³ - x² +1) / **-(x² +x +1)-** = x³ -x² +1*
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It is a true art to take a math problem and be able to come up with 3 bad methods/explanations. Impressive, as always.
Ahaha! Thanks for the kind words!!!
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My bro doesn't heed on useless people whose work is to find only problems in other people 😊😊