(i)(x^2+x-2)^4+(2x+1)^4=(x^2+x-1)^4\\\\ (ii)(x^2+3x+2)(x^2+7x+12)+(x^2+5x-6)=0
62(x+1)(x+2)(x+3)(x+4)+(x-1)(x+6)=0
t=x^2+5x
(t+4)(t+6)+(t-6)=0
t^2+11t+18=0
t=-2 or t=-9
1708Yes Nishant Sir You are saying Right.
\textbf{for first question}::(i)\\\\ \mathbf{(x^2-x-2)^4+(2x-1)^4=(x^2+x-1)^4}\\\\ $(There was an Typo error.)
341Did you mean real solutions? If so,
The equation is of the form a^4+b^4 = (a+b)^4
Setting y = \frac{a}{b} we have
y^4+1 = (y+1)^4 \Rightarrow y(2y^2+3y+2) =0
Since the quadratic factor is positive for all real y, we must have y=0
i.e. a=0.
Arguing similarly, b=0 is also a solution
Hence the solutions correspond to those of x^2-x-2 = 0 and
that of 2x+1=0
1708Yes Sir I am searching for real solution.
So \mathbf{x=-1,-\frac{1}{2},2}
Thanks hsbhatt Sir.
62oh now i know why i was banging my head over this one for 20 minutes :P