The equation $16 x^4+16 x^3-4 x-1=0$ has a multiple root. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation, then $\frac{1}{\alpha^4}+\frac{1}{\beta^4}+\frac{1}{\gamma^4}+\frac{1}{\delta^4}=$

Given: $16 x^4+16 x^3-4 x-1=0=f(x)$ has a multiple root.
Let $\alpha=\beta$ be the multiple root. Since $\alpha$ is root of $f(x) \quad \therefore f(\alpha)=0$ $\Rightarrow 16 \alpha^4+16 \alpha^3-4 \alpha-1=0$ ...(i) As $\alpha$ is multiple root $\Rightarrow f^{\prime}(\alpha)=0$ Now, $f^{\prime}(x)=64 x^3+48 x^2-4 \Rightarrow f^{\prime}(\alpha)=0$ $\Rightarrow 64 \alpha^3+48 \alpha^2-4=0 \Rightarrow 16 \alpha^3+12 \alpha^2-1=0$ ...(ii) Subtracting (ii) from (i), $16 \alpha^4-12 \alpha^2+1-4 \alpha-1=0$ $\Rightarrow 16 \alpha^4-12 \alpha^2-4 \alpha=0 \Rightarrow 4 \alpha\left(4 \alpha^3-3 \alpha-1\right)=0$ $\alpha \neq 0$ as it doesn't satisfy given equation $\Rightarrow \cdot 4 \alpha^3-3 \alpha-1=0 \Rightarrow(\alpha-1)(2 \alpha+1)^2=0$ $\alpha \neq 1$ as it doesn't satisfy given equation $\Rightarrow 2 \alpha+1=0 \Rightarrow \alpha=\frac{-1}{2}$ So, $\alpha=\frac{-1}{2}$ is a root of the given equation By using synthetic division $\Rightarrow g(x)=16 x^3+8 x^2-4 x-2$ Since, $\alpha=\frac{-1}{2}$ is multiple root By using synthetic division again $\Rightarrow h(x)=16 x^2-4=0$ $\Rightarrow x^2=\frac{4}{16}=\frac{1}{4} \Rightarrow x= \pm \frac{1}{2}$
Hence, $\alpha=\frac{-1}{2}, \beta=\frac{-1}{2}, \gamma=\frac{-1}{2}, \delta=\frac{1}{2}$ So, $\frac{1}{\alpha^4}+\frac{1}{\beta^4}+\frac{1}{\gamma^4}+\frac{1}{\delta^4}=2^4+2^4+2^4+2^4$ $=16+16+16+16=64 .$