The degree of the differential equation
\(1+\left(\frac{d y}{d x}\right)^2+\left(\frac{d^2 y}{d x^2}\right)^2=\sqrt[3]{\frac{d^2 y}{d x^2}+1}\) is

Given,
\(1+\left(\frac{d y}{d x}\right)^2+\left(\frac{d^2 y}{d x^2}\right)^2=\sqrt[3]{\frac{d^2 y}{d x^2}+1}\)
On cubic both sides, we get
\(\begin{aligned} & {\left[1+\left(\frac{d y}{d x}\right)^2+\left(\frac{d^2 y}{d x^2}\right)^2\right]^3=\left(\frac{d^2 y}{d x^2}+1\right) } \\ \Rightarrow \quad & {\left[1+\left(\frac{d y}{d x}\right)^2\right]^3+\left(\frac{d^2 y}{d x^2}\right)^6+3\left[1+\left(\frac{d y}{d x}\right)^2\right] \cdot \frac{d^2 y}{d x^2} } \\ & {\left[1+\left(\frac{d y}{d x}\right)^2+\left(\frac{d^2 y}{d x^2}\right)^2\right]=\frac{d^2 y}{d x^2}+1 }\end{aligned}\)
So, degree of the differential equation is 6 .