• Sep 16th 2009, 11:14 PM
jashansinghal
what will be the ultimate result
(a+b+c)(-a+b+c)(a-b+c)(a+b-c)
• Sep 17th 2009, 12:32 AM
Hello jashansinghal
Quote:

Originally Posted by jashansinghal
what will be the ultimate result
(a+b+c)(-a+b+c)(a-b+c)(a+b-c)

If we group the brackets in pairs we can write:

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

$=\Big( ([b+c]+a)([b+c]-a)\Big)\Big((a-[b-c])(a+[b-c])\Big)$

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

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

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

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