# A factorisation question

• October 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)
• October 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: . $(b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)$
$b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4$

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer: . ${\color{blue}(a-b)(b-c)(c-a)(a+b+c)}$

• October 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer: . ${\color{blue}(a-b)(b-c)(c-a)(a+b+c)}$

Wow!
• October 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 $a = b$ in the expression, we get:

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

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

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

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

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

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

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

Or did I cheat a bit?

• October 1st 2009, 08:37 AM
pacman
great work soroban and grandad, whew!

http://www.mathhelpforum.com/math-he...5ea190ad-1.gif = -(a - b)(a - c)(b - c)(a + b + c).

as was observed by grandad, a = b = c. But if you let a = -(b + c); the expression vanishes too.

Then from a = -(b + c), we have (a + b + c) as a factor also.