Question

1^4+2007^4+2008^4/1^2+2007^2+2008^2=2007^2+p then p=

Answer 1

Answer:

You can calculate (t-1)^4 by two methods

one by binomial method and other by normal process.

(x-1)^n=(1-x)^n= 1- nx+{n(n-1)/2!}x^2 -{n(n-1)(n-2)/3! }*x^3 +....so on.

normal process would be little bit lengthy,but easier if you haven't tasted higher algebra yet.

(t-1)^4= {(t-1)^2}^2=(t^2-2t+1)^2=t^4+4t^2+1-4t^3-4t+2t^2

=t^4-4*t^3+6t^2-4t+1

Step-by-step explanation:

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