A factorisation question

• Oct 1st 2009, 03:03 AM
anshulbshah
A factorisation question
Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)
• Oct 1st 2009, 05:52 AM
Soroban
Hello, anshulbshah!

I expanded it, then repeatedly factored "by grouping".
If there is a more direct way, I hope someone finds it.

Quote:

Factor: .\$\displaystyle (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)\$
\$\displaystyle b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4\$

. . \$\displaystyle = \;ab^3 - ac^3 - a^3b + a^3c - b^3c + nbc^3\$

. . \$\displaystyle = \;a(b^3-c^3) - a^3(b-c) - bc(b^2-c^2)\$

. . \$\displaystyle =\; a{\color{red}(b-c)}(b^2+bc+c^2) - a^3{\color{red}(b-c)} - bc{\color{red}(b-c)}(b+c) \$

. . \$\displaystyle =\; {\color{red}(b-c)}\bigg[a(b^2+bc+c^2) - a^3 - bc(b+c)\bigg] \$

. . \$\displaystyle =\; (b-c)\bigg[ab^2 + abc + ac^2 - a^2 - b^2c - bc^2\bigg]\$

. . \$\displaystyle =\; (b-c)\bigg[ac^2 - a^3 - b^2c + ab^2 - bc^2 + abc\bigg]\$

. . \$\displaystyle =\; (b-c)\bigg[a(c^2-a^2) - b^2(c-a) - bc(c-a)\bigg] \$

. . \$\displaystyle = \;(b-c)\bigg[a{\color{red}(c-a)}(c+a) - b^2{\color{red}(c-a)} - bc{\color{red}(c-a)}\bigg] \$

. . \$\displaystyle =\; (b-c){\color{red}(c-a)}\bigg[a(c+a) - b^2 - bc\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a)\bigg[ac + a^2 - b^2 - bc\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a)\bigg[a^2-b^2 + ac - bc\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a)\bigg[{\color{red}(a-b)}(a+b) + c{\color{red}(a-b)}\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a){\color{red}(a-b)}\bigg[a+b+c\bigg]\$

• Oct 1st 2009, 06:13 AM
masters
Quote:

Originally Posted by Soroban
Hello, anshulbshah!

I expanded it, then repeatedly factored "by grouping".
If there is a more direct way, I hope someone finds it.

\$\displaystyle b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4\$

. . \$\displaystyle = \;ab^3 - ac^3 - a^3b + a^3c - b^3c + nbc^3\$

. . \$\displaystyle = \;a(b^3-c^3) - a^3(b-c) - bc(b^2-c^2)\$

. . \$\displaystyle =\; a{\color{red}(b-c)}(b^2+bc+c^2) - a^3{\color{red}(b-c)} - bc{\color{red}(b-c)}(b+c) \$

. . \$\displaystyle =\; {\color{red}(b-c)}\bigg[a(b^2+bc+c^2) - a^3 - bc(b+c)\bigg] \$

. . \$\displaystyle =\; (b-c)\bigg[ab^2 + abc + ac^2 - a^2 - b^2c - bc^2\bigg]\$

. . \$\displaystyle =\; (b-c)\bigg[ac^2 - a^3 - b^2c + ab^2 - bc^2 + abc\bigg]\$

. . \$\displaystyle =\; (b-c)\bigg[a(c^2-a^2) - b^2(c-a) - bc(c-a)\bigg] \$

. . \$\displaystyle = \;(b-c)\bigg[a{\color{red}(c-a)}(c+a) - b^2{\color{red}(c-a)} - bc{\color{red}(c-a)}\bigg] \$

. . \$\displaystyle =\; (b-c){\color{red}(c-a)}\bigg[a(c+a) - b^2 - bc\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a)\bigg[ac + a^2 - b^2 - bc\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a)\bigg[a^2-b^2 + ac - bc\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a)\bigg[{\color{red}(a-b)}(a+b) + c{\color{red}(a-b)}\bigg]\$

. . \$\displaystyle =\; (b-c)(c-a){\color{red}(a-b)}\bigg[a+b+c\bigg]\$

Wow!
• Oct 1st 2009, 06:55 AM
Hello anshulbshah
Quote:

Originally Posted by anshulbshah
Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)

Having seen Soroban's answer, I suppose this is being wise after the event, but here's a quicker method, if you spot it!

If we put \$\displaystyle a = b\$ in the expression, we get:

\$\displaystyle (b-c)(b^3+c^3) + (c-b)(c^3+b^3) + 0 = 0\$

so \$\displaystyle (a-b)\$ is a factor.

Similarly \$\displaystyle (b-c)\$ and \$\displaystyle (c-a)\$ are also factors.

Since the expression is of degree \$\displaystyle 4\$ in \$\displaystyle a,b,c\$, the remaining factor is linear, and by symmetry is therefore \$\displaystyle k(a+b+c)\$ for some constant \$\displaystyle k\$.

So \$\displaystyle (b-c)(b^3+c^3)+(c-a)(c^3+a^3)+(a-b)(a^3+b^3) = k(a-b)(b-c)(c-a)(a+b+c)\$

Compare coefficients of, say, \$\displaystyle bc^3: 1 = k\$ (taking the \$\displaystyle -b\$ from the first factor, the \$\displaystyle -c\$ from the second, the \$\displaystyle c\$ from the third and the \$\displaystyle c\$ from the fourth)

\$\displaystyle \Rightarrow (b-c)(b^3+c^3)+(c-a)(c^3+a^3)+(a-b)(a^3+b^3) = (a-b)(b-c)(c-a)(a+b+c)\$

Or did I cheat a bit?