Please Factorise (1 Viewer)

math man · Feb 11, 2012

my final answer is $24abc$ but i did it really fast so
i might of made an error. All I did was factorise using sum and difference of 2 cubes

Drongoski · Feb 11, 2012

math man said:
my final answer is $8abc$ but i did it really fast so
i might of made an error. All I did was factorise using sum and difference of 2 cubes

For Q1 or Q2?

math man · Feb 11, 2012

Shadowdude · Feb 11, 2012

WolframAlpha is your friend.

The first one apparently gives: $(a^{2} - b^{2})(a^{2}-c^{2})(b^{2}-c^{2})$

Second is apparently $24abc$

math man · Feb 11, 2012

yeh it is 24abc i forgot the extra 4

kaz1 · Feb 11, 2012

was bored, couldn't do no.1

no. 2 is 24abc

Drongoski · Feb 11, 2012

math man said:
yeh it is 24abc i forgot the extra 4

Not bad. How about Q1? Wolfram is a Genius. Shadow has given the right answers. But how do you do them?

math man · Feb 11, 2012

i cant be bothered typing it so long, but for 2 just apply difference of cubes on first two terms
and sum of two cubes on second two terms and after that complete the square and keep
factorising

Drongoski · Feb 12, 2012

kaz1 said:
was bored, couldn't do no.1

no. 2 is 24abc

Good effort anyway.

4025808 · Feb 12, 2012

for Q2, expand it out, then + and - a^2*b^2*c^2. From there, rearrange and factorize. It really helps working it out.

Manipulation techniques really come in handy.

math man · Feb 12, 2012

4025808 said:
for Q2, expand it out, then + and - a^2*b^2*c^2. From there, rearrange and factorize. It really helps working it out.

Manipulation techniques really come in handy.

You don't have to expand it, jus factorise immediately
As difference and sum of cubes instead

SpiralFlex · Feb 12, 2012

No one wanted to do Q1.

$a^4(b^2-c^2) + b^4(c^2-a^2) + c^4(a^2-b^2)$

$=a^4b^2-a^4c^2+b^4c^2-b^4a^2+c^4a^2-c^4b^2$

$=a^4b^2-a^2b^4-c^2a^4+a^2b^2c^2-a^2b^2c^2+b^4c^2+c^4a^2-c^4b^2$

$=(b^2a^2-b^4-c^2a^2+c^2b^2)(a^2-c^2)$

$=[b^2(a^2-b^2)-c^2(a^2-b^2)](a^2-c^2)$

$=(b^2-c^2)(a^2-c^2)(a^2-b^2)$

$=(b-c)(b+c)(a-c)(a+c)(a-b)(a+b)$

nightweaver066 · Feb 12, 2012

This thread made me want to learn cyclic expansion & factorisation. Still new, it's hard to find information about it online..

Unsure if the way i set out the solution is correct or not..

$$2) Let $ f(a, b, c) = (a + b + c)^3 - (b + c - a)^3 - (c + a - b)^3 - (a + b - c)^3$

$f(b, a, c) = (a + b + c)^3 - (a + c - b)^3 - (c + b - a)^3 - (a + b - c)^3$

$= f(a, b, c)$

$\therefore $ It is cyclic$$

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

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

$= 0$

$\therefore $ a is a factor by factor theorem. Similarly, as it is a cyclic expression, b and c are also factors$$

$deg[f(a, b, c)] = 3$

$\therefore Kabc \equiv (a + b + c)^3 - (a + c - b)^3 - (c + b -a)^3 - (a + b - c)^3 $ where K is a real number$

$Let a = 1, b = 1, c = 1$

$K = 27 - 1 - 1 - 1$

$= 24$

$\therefore f(a, b, c) = 24abc$

4025808 · Feb 12, 2012

math man said:
You don't have to expand it, jus factorise immediately
As difference and sum of cubes instead

oh woops my bad I meant Q1

Trebla · Feb 12, 2012

Drongoski said:
$1) a^4(b^2-c^2) + b^4(c^2-a^2) + c^4(a^2-b^2)$

$a^4(b^2-c^2) + b^4(c^2-a^2) + c^4(a^2-b^2)\\\\=a^4b^2-a^4c^2+b^4c^2-a^2b^4+c^4(a^2-b^2)\\\\=a^2b^2(a^2-b^2)-c^2(a^4-b^4)+c^4(a^2-b^2)\\\\=a^2b^2(a^2-b^2)-c^2(a^2-b^2)(a^2+b^2)+c^4(a^2-b^2)\\\\=(a^2-b^2)(a^2b^2-c^2(a^2+b^2)+c^4)\\\\=(a^2-b^2)(a^2b^2-a^2c^2-b^2c^2+c^4)\\\\=(a^2-b^2)(a^2(b^2-c^2)-c^2(b^2-c^2))\\\\=(a^2-b^2)(b^2-c^2)(a^2-c^2)\\\\=(a+b)(a-b)(b+c)(b-c)(a+c)(a-c)$

Trebla · Feb 12, 2012

nightweaver066 said:
This thread made me want to learn cyclic expansion & factorisation. Still new, it's hard to find information about it online..

Unsure if the way i set out the solution is correct or not..

$$2) Let $ f(a, b, c) = (a + b + c)^3 - (b + c - a)^3 - (c + a - b)^3 - (a + b - c)^3$

$f(b, a, c) = (a + b + c)^3 - (a + c - b)^3 - (c + b - a)^3 - (a + b - c)^3$

$= f(a, b, c)$

$\therefore $ It is cyclic$$

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

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

$= 0$

$\therefore $ a is a factor by factor theorem. Similarly, as it is a cyclic expression, b and c are also factors$$

$deg[f(a, b, c)] = 3$

$\therefore Kabc \equiv (a + b + c)^3 - (a + c - b)^3 - (c + b -a)^3 - (a + b - c)^3 $ where K is a real number$

$Let a = 1, b = 1, c = 1$

$K = 27 - 1 - 1 - 1$

$= 24$

$\therefore f(a, b, c) = 24abc$

I think you have to show it is equal for all variations to prove it is a cyclic expression
i.e. prove f(a, b, c) = f(a, c, b) = f(b, a, c) = f(b, c, a) = f(c, a, b) = f(c, b, a)

gurmies · Feb 12, 2012

Trebla said:
I think you have to show it is equal for all variations to prove it is a cyclic expression
i.e. prove f(a, b, c) = f(a, c, b) = f(b, a, c) = f(b, c, a) = f(c, a, b) = f(c, b, a)

Typically you do, but the above is clearly cyclic.

Please Factorise (1 Viewer)

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