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caylus
Hello,

(u+v)^4= u^4+4^3v+6u²v²+4uv^3+v^4

with u=2x² and v=y²

(2x²+y²)^4=
16x^8 + 32x^6*y² +24x^4y^4 +8x²y^6 + y^8

olemakpadu
(2x² + y²)⁴

By Binomial Expansion:

⁴C₀(2x²)⁴(y²)⁰  +  ⁴C₁(2x²)³(y²)¹ +  ⁴C₂(2x²)²(y²)² +  ⁴C₃(2x²)¹(y²)³  + ⁴C₄(2x²)⁰(y²)⁴

From Pascal's Triangle:

Note the C is the values from Combinations.

For power 4, the  factors are:

1 ,  4,  6, 4, 1

1*(2x²)⁴(y²)⁰  +  4*(2x²)³(y²)¹ +  6*(2x²)²(y²)² +  4*(2x²)¹(y²)³  + 1*(2x²)⁰(y²)⁴

Note if it raised to power zero = 1.

(2⁴x² ˣ ⁴)  + 4*2³x² ˣ ³y² ˣ ¹  + 6*2²x² ˣ ² y² ˣ ² +  4*2x² ˣ ¹y² ˣ ³  +  y² ˣ ⁴

16x⁸ + 32x⁶y² + 24x⁴y⁴ + 8x²y⁶ + y⁸

I hope this helped.

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