If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$

Question

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?

Answer 1

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 $1+p^2+p^4=(1+p^2)^2-(p)^2=?$

Answer 2

Hint : $(2^{\frac{1}{4}}-1) \cdot (2^\frac{1}{2} + 2^\frac{1}{4} +1) = 2^\frac{3}{4} -1$

Hint : $7 = 2^3-1$