The equation left(x^4+1right)=frac{1}{a}(x+1) | vedclass

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(B) Given equation is $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$. Multiplying by $a$,we get $a(x^4+1) = (x+1)^4$. Expanding the right side: $a(x^4+1) = x

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for all $a \in R-\{1\}$

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(B) Given equation is $\left(x^4+1ight)=\frac{1}{a}(x+1)^4$. Multiplying by $a$,we get $a(x^4+1) = (x+1)^4$. Expanding the right side: $a(x^4+1) = x^4+4x^3+6x^2+4x+1$. Rearranging the terms: $(a-1)x^4 - 4x^3 - 6x^2 - 4x + (a-1) = 0$. For an equation to be a reciprocal equation,the coefficients of $x^k$ and $x^{n-k}$ must be equal (or proportional). Here,the coefficient of $x^4$ is $(a-1)$ and the constant term is $(a-1)$. For the equation to be of degree $4$,the coefficient of $x^4$ must not be zero,so $a-1 
eq 0$,which means $a 
eq 1$. Thus,the equation is a reciprocal equation for all $a \in R - \{1\}$.

The equation $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$ is a reciprocal equation:

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