# Just deleted my working

• Jun 12th 2009, 06:52 AM
Joel
Just deleted my working
I accidentally just deleted some working and cannot get it back.

Can you help me with this please.

Factor fully: ( a² - b² - c² )² - 4b²c²

= (a²-b²-c²)² - (2bc)²
= (a² - b² - c² - 2bc)(a² - b² - c² + 2bc)
= (a² - (b + c)²)(a² - (b – c)²)
=
=
= (a-b-c)(a+b+c)(a-b+c)(a+b-c)
• Jun 12th 2009, 06:58 AM
alexmahone
Quote:

Originally Posted by Joel
I accidentally just deleted some working and cannot get it back.

Can you help me with this please.

Factor fully: ( a² - b² - c² )² - 4b²c²

= (a²-b²-c²)² - (2bc)²
= (a² - b² - c² - 2bc)(a² - b² - c² + 2bc)
= (a² - (b + c)²)(a² - (b – c)²)
= (a-(b+c))(a+(b+c))(a-(b-c))(a+(b-c))
= (a-b-c)(a+b+c)(a-b+c)(a+b-c)

Missing working has been colored in red.
• Jun 12th 2009, 06:59 AM
Amer
Quote:

Originally Posted by Joel
I accidentally just deleted some working and cannot get it back.

Can you help me with this please.

Factor fully: ( a² - b² - c² )² - 4b²c²

= (a²-b²-c²)² - (2bc)²
= (a² - b² - c² - 2bc)(a² - b² - c² + 2bc)
= (a² - (b + c)²)(a² - (b – c)²)
=
=
= (a-b-c)(a+b+c)(a-b+c)(a+b-c)

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

$\left[(a-(b+c))(a+(b+c))\right]\left[(a-(b-c))(a+(b-c))\right]$

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

someone is faster than me .......
• Jun 12th 2009, 07:15 AM
Joel
Thanks Guys