Question

(3ײ+5y²)³= (2y³-×5)²=

Answer 1

First one:
(3x^2+5y^2)^3
Step 1: Use the binomial theorem to expand
(3x^2)+3(3x^2)^2(5y^2)+3(3x^2)(5y^2)2+(5y^2)^3
Step 2: Simplify each term
27(x^2)^3+3(3x^2)^2(5y^2)+3(3x^2)(5y^2)^2+(5y^2)^3
Step 3: Multiply the exponents
27x^6+27x^4(5y^2)+3(3x^2)(5y^2)^2+(5y^2)^3
27x^5+135x^4y^2+3(3x^2)(5y^2)^2+(5y^2)^3
Keep going to get the final answer:
27x^6+135x^4y^2+225x^2y^4+125y^6

2nd question: (2y^3-5x)^2
Do the same thing...step by step
Answer: 4y^6-20xy^3+25x^2