If “

     

My initial thought is that if $a$, $b$ và $c$ are $1$ or $-1$, then the polynomial evaluates to lớn $0.$ So, maybe two of the factors will be $(a + b + c - 3)$ and $(a + b + c + 3)$. An alternative option that combines these two might be $a^2 + b^2 + c^2 - 3$.

Is the thought process correct here, & would trial và error be a good way lớn decide between the linear and the quadratic options I described above?As you can tell, I am largely doing guess work here. Is there a more systematic way of deciding what terms to địa chỉ and subtract in orders khổng lồ factor the polynomial?

Note: The factoring need not be done all the way lớn linear factors. All that is needed is a hàng hóa of polynomials.

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I suggests that you use $(a+b)^3=a^3+b^3+3ab(a+b)Rightarrow a^3+b^3=(a+b)^3-3ab(a+b)$ instead, you will need to lớn use it twice lượt thích this:

$a^3+b^3+c^3-3abc$

$=(a+b)^3+c^3-3ab(a+b)-3abc$

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

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

$=(a+b+c)^3-(a+b+c)(3ab+3bc+3ca)$

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

$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$


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Hint: divide $$a^3+b^3+c^3-3abc$$ by $a+b+c$ the result is given by $$left( c+a+b ight) left( a^2-ab-ca+b^2-bc+c^2 ight) $$


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Spivak, Ch 2 Prologue, Problem 12d: is there a systematic way lớn find two irrational numbers whose sum and multiplication result in rational number?
For what constants $a,b,c,d,e,f,gingoodsmart.com.vnbfZ$ is $(ax^2+bx+c)(fx+g)-(dx+e)=0$ solvable via the *Rational Root Thm.*
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