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Should I use \(a^3 + b^3 = (a + b)(a^2 - ab+b^2)\)?
Or am I complicating it?
The factorization is not that easy to find

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Other answers:

Yeah, but Wolfram does give a simple solution.
Why simple? Complex numbers join the party and you can't even find a real root to get started...
Yeah, but the solution is simple. :-P This is 9th grade math... I don't know why I'm not able to do this one
Gimme a minute...
\[\begin{array}{l} 9{x^4} + 9{x^3} + 8{x^2} + 9x + 9 = 0 \\ 9({x^4} + {x^3} + 2{x^2} + x + 1) = 10{x^2} \\ 9{({x^2} + 1)^2} + 9x({x^2} + 1) - 10{x^2} = 0 \\ {\left( {3({x^2} + 1)} \right)^2} + \left( {3({x^2} + 1)} \right)(3x) - 10{x^2} = 0 \\ {\left( {3({x^2} + 1)} \right)^2} - \left( {3({x^2} + 1)} \right)(2x) + \left( {3({x^2} + 1)} \right)(5x) - (2x)(5x) = 0 \\ \end{array}\] Actually 4 mins already, and I left out the last step :P
Ha, but how did you choose to add \(10x^2\) to both sides? Doesn't it seem arbitrary?
No it does not. At first glance I was going to add x^2 only, but then to complete the square I have to add another 9x^2
How do you complete the square in a polynomial with degree 3 or higher? :-O
Oh lol
Got it
\[(3x)^2 + (3)^2\]has the middle term \(2(3x)^2 (3)^2 = 18x^2\). Thanks :-)
just pick the dominant term with even degree and add/subtract until you have a complete square
KUDOS FOR DRAWAR , GREAT WORK MR. DRAWAR :)
Thanks for that @mathslover :) Oh kids have to study so many things nowadays...
@drawar Can you send me the link where you learnt how to do this?
Sorry I don't have any link to share here, just do more practice and you'll eventually get the hang of it!
OK :-|
Just out of curiosity, are you a 9th grader?
Yes.
Just started 9th grade.
Nice, all the best with your study then!
Thanks!

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