Solution ①: The original equation becomes: 2-x²√2-x³-x²=0 Let a=√2, then a²=2∴a²-x²a-x³-x²=0△=x⁴+4x³+4x²= (x²+2x) ²a=x²+x or a=-x when a=x²+x, x²+x-√2=0, x=[-1+√(1+4√2)]/2 or x=[-1-√(1+4√2)]/2 when a=-

Solution ①: The original equation becomes:

2-x²√2-x³-x²=0

Let a=√2, then a²=2

∴a²-x²a-x³-x²=0

△=x⁴+4x³+4x²=（x²+2x）²

a=x²+x or a=-x

When a=x²+x, x²+x-√2=0, x=[-1+√(1+4√2)]/2 or x=[-1-√(1+4√2)]/2

When a=-x, √2=-x, x=-√2

∴The solution of the original equation is: ∴x1=- √2.

∴(x+√2)(x²-√2x+2)+(1+√2)x²-2(1+√2)=0

∴(x+√2)(x²-√2x+2)+(1+√2)(x+√2)(x-√2)=0

∴(x+√2)(x²-√2x+2+x-√2+√2x-2)=0

∴(x+√2)(x²+x-√2)=0

∴The solution of the original equation is: x1=-√2, x2=[-1+√(1+4√2)]/2 , x3=[-1-√(1+4√2)]/2

Solution ③: Trial root method, x=-√2 is the root of the original equation

∴The original equation can become: (x+√2) (x²+x-√2)=0

∴x1=-√2, x2=[-1+√(1+4√2)]/2, x3=[-1-√(1+4√2)]/2

∴The solution of the original equation is:∴x1=-√2, x2=[-1+√(1+4√2)]/2, x3=[-1-√(1+4√2)]/2