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A step by step solution would be preferred for the following question :

Find the values of the constants in the following identitity $x^4+4/x^4 = (x^2-A/X^2)^2+B$.

so far I managed to substitute $x$ for $1$ to get :


However I'm not sure how to proceed further.

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up vote 1 down vote accepted

Let's find the coefficients without using the Heaviside cover-up method (that is, without substituting values for $x$). First, we convert to polynomials by squaring and clearing fractions: $$ \begin{align*} x^4+\dfrac{4}{x^4} &= \left(x^2-\dfrac{A}{x^2}\right)^2+B \\ x^4+\dfrac{4}{x^4} &= \left(x^4-2A+\dfrac{A^2}{x^4}\right)+B \\ \dfrac{4}{x^4} &= -2A+B+\dfrac{A^2}{x^4} \\ 4 &= (-2A+B)x^4+A^2 \\ \end{align*} $$ Next, we compare coefficients. That is, we equate the coefficients of each power of $x$ for the polynomials for each side. In this case, we need only compare the coefficients of $x^4$ and $x^0$ (the constants): $$ \begin{align*} x^0&: \quad 4=A^2 \implies A=\pm2\\ x^4&: \quad 0=-2A+B \implies B=2A=2(\pm2)=\pm4\\ \end{align*} $$

So there are two possibilities. Either $\boxed{A=2,B=4}$ or $\boxed{A=-2,B=-4}$.

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it would be really helpful if you could explain further why x0 and x^4 are chosen. I'm not quite sure of the procedure for comparing coefficents. for example if I've got x^4+x^2+x+1 = (x^4-x^2+Ax^2-A)+Bx+c. I should remove the brackets first and then look to compare only those powers of x with coefficents -x^2+Ax^2-A? – peter_gent Jun 12 '13 at 0:04
@peter_gent That's correct. In your example, we have: $$ x^4+x^2+x+1=x^4+(A-1)x^2+Bx+(C-A) $$ Comparing the coefficients of $x^4$ or $x^3$ isn't very helpful (it yields $1=1$ or $0=0$, which is true, but doesn't help us solve for $A,B,C$). Thus, we compare the coefficients of $x^2$ (which yields $1=A-1$) and $x^1$ (which yields $1=B$) and $x^0$ (which yields $1=C-A$). – Adriano Jun 12 '13 at 4:34

Your calculation is not right as $$\left(x^2-\frac A{x^2}\right)^2+B=x^4-2A+\frac{A^2}{X^4}+B$$

So comparing the coefficients of $\frac1{x^4}$, $A^2=4$

and comparing the constants $-2A+B=0$

Can you take it from here?

Alternatively use $a^2+b^2=(a+ b)^2-2ab$ or $a^2+b^2=(a-b)^2+2ab$

Here $a^2=x^4\implies a=\pm x^2$ and $ b^2=\frac4{x^4}\implies b=\pm \frac2{x^2}$

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does this mean you also have to ^2 the rhs ? – peter_gent Jun 11 '13 at 16:08
@peter_gent, yes. Please find the alternate method – lab bhattacharjee Jun 11 '13 at 16:15
2 secs I'll give it a go. thanks :) – peter_gent Jun 11 '13 at 16:17

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