<s>[INST] If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$ 
If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$ .

What I Tried: I wrote :-
$$\rightarrow \frac{7}{2^{1/2} + 2^{1/4} + 1} = \frac{2^2 + 2 + 1}{2^{1/2} + 2^{1/4} + 1}$$
Then I rationalized the denominator to get :-
$$\rightarrow\frac{(2^2 + 2 + 1)(2^{1/2} + 2^{1/4} - 1)}{(2^{1/2} + 2^{1/4} + 1)(2^{1/2} + 2^{1/4} -1)}$$
$$\rightarrow \frac{2^{5/2}+2^{9/4}-2^2+2^{3/2}+2^{5/4}-2+2^{1/2}+2^{1/4}-1}{(2^{1/2} + 2^{1/4})^2 - 1}$$
$$\rightarrow  \frac{2^{5/2}+2^{9/4}-2^2+2^{3/2}+2^{5/4}-2+2^{1/2}+2^{1/4}-1}{( 2^{5/4}+2^{1/2} + 1)}$$
I am stuck here. I probably cannot proceed further unless I find a way to factorise the numerator, but I have not find a way to do that for now. Can anyone help me with this problem?
 [/INST] Let $2^{1/4}=p$
So, we need
$$\dfrac{1+p^4+p^8}{1+p+p^2}$$
Now $1+p^4+p^8=(1+p^4)^2-(p^2)^2=?$
and $1+p^2+p^4=(1+p^2)^2-(p)^2=?$
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