# 3abc

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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.

Bạn đang xem: 3abc 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 . Thanks for contributing an answer to xechieuve.com.vnematics Stack Exchange!

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Asking for help, clarification, or responding to lớn other answers.Making statements based on opinion; back them up with references or personal experience.

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