1but sir,for me coefficient of x^2 is coming to d answer u got in coefficient of x........
Δ=
1+x (1+x)a (1+x)bc
1+x (1+x)b (1+x)ca
1+x (1+x)c (1+x)ab
find coeff. of x and x2.
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7 Answers
62(1+x)\left\{(1+x)^{b^2a}-(1+x)^{c^2a}+(1+x)^{c^2b}-(1+x)^{a^2b}+(1+x)^{a^2c}-(1+x)^{b^2c}\right\}=x\left\{(1+x)^{b^2a}-(1+x)^{c^2a}+(1+x)^{c^2b}-(1+x)^{a^2b}+(1+x)^{a^2c}-(1+x)^{b^2c}\right\}+\left\{(1+x)^{b^2a}-(1+x)^{c^2a}+(1+x)^{c^2b}-(1+x)^{a^2b}+(1+x)^{a^2c}-(1+x)^{b^2c}\right\}
There should be a simpler method.. but right now i cant think of any! :(
1take (1+x)^abc common from the third row
u get
Δ=
1+x (1+x)^a (1+x)^1/a
1+x (1+x)^b (1+x)^1/b
1+x (1+x)^c (1+x)^1/c
now put 1+x=t
and solve
1according to arshad,i hav reached to
=> t^abc [ t^ab-bc-a2+2ac - t^ac-bc-a2+2ab ]
can ne1 help me after dis???
62after my 2nd post..
coeff of x seems to be obvious?
b2a+c2b+a2c-a2b-b2c-c2a
=(a-b)(b-c)(c-a)
and for x^2 it will be
seems to be a bit more clumsy.. :(
162damn.. i have kept this one posted as teh most stupid wrong answer and no one ever saw what i did :( :( :(
There is a stupid stupid stupid mistake in what i have done :D
let me rewrite it...
(1+x)\left\{(1+x)^{b+ba}-(1+x)^{c+ca}+(1+x)^{c+cb}-(1+x)^{a+ab}+(1+x)^{a+ac}-(1+x)^{b+bc}\right\}
Now can you guys solve it?