3ABC

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

Is the thought process correct here, and would trial và error be a good way to 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 add and subtract in orders to lớn factor the polynomial?

Note: The factoring need not be done all the way lớn linear factors. All that is needed is a sản phẩm 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 lớn use it twice like this:

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

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

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

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

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

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

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

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

Factor $a^3+b^3+c^3-3abc$ khổng lồ a product of polynomials.

Think when $a=b=c$.

$$a^3+b^3+c^3-3abc=a^3+a^3+a^3-3a^3=0.$$

This only fits when $a=b=c$, but not when $a=b$ or $b=c$ or $c=a$.

So, you can think about $(a-b)^2+(b-c)^2+(c-a)^2$, which is $0$ if and only if $a=b=c$.

Therefore, you can reason $a^2+b^2+c^2-ab-bc-ca$ as the factor of $a^3+b^3+c^3-3abc$.

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Also, think when $a=-b-c$.

$$a^3+b^3+c^3-3abc=(-b-c)^3+b^3+c^3-3(-b-c)bc \ =-b^3-3b^2c-3bc^2-c^3+b^3+c^3+3(b+c)bc=0.$$

ISW, you can reason $a+b+c$ as the factor of $a^3+b^3+c^3-3abc.$

Since the degree of $a^2+b^2+c^2-ab-bc-ca$ is $2$, while $a+b+c$ has $1$, You can reason $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$. Checking this, you can find that you got the right one.

Since the given polynomial is homogeneous và symmetrical w.r.t. A,b,c there can be only one linear factor a+b+c. You can substitute -(b+c) for a và prove that the result is zero & verify that a factor is a+b+c. Since the polynomial is of 3rd degree the remaining factor should be of the form A(a² + b² + c²) +B(ab+bc+ca). Now by equating the coefficients or by substituting values for a,b,c it can be obtained A =1 , B = -1 .

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