You used combination method I want you to use pascal

The 6th row of Pascal's Triangle (starting with row 0) provides the coefficients:

$$1, 6, 15, 20, 15, 6, 1$$

The general form is $C_k a^{n-k} b^k$, where $C_k$ is the coefficient from Pascal's Triangle.

• Term 1 ($k=0$): Coefficient is 1.

  $$1 \cdot (3)^{6-0} \cdot (5x)^0 = 1 \cdot 3^6 \cdot 1 = 1 \cdot 729 \cdot 1 = 729$$

• Term 2 ($k=1$): Coefficient is 6.

  $$6 \cdot (3)^{6-1} \cdot (5x)^1 = 6 \cdot 3^5 \cdot 5x = 6 \cdot 243 \cdot 5x = 6 \cdot 1215x = 7290x$$

• Term 3 ($k=2$): Coefficient is 15.

  $$15 \cdot (3)^{6-2} \cdot (5x)^2 = 15 \cdot 3^4 \cdot (25x^2) = 15 \cdot 81 \cdot 25x^2 = 15 \cdot 2025x^2 = 30375x^2$$

• Term 4 ($k=3$): Coefficient is 20.

  $$20 \cdot (3)^{6-3} \cdot (5x)^3 = 20 \cdot 3^3 \cdot (125x^3) = 20 \cdot 27 \cdot 125x^3 = 20 \cdot 3375x^3 = 67500x^3$$

• Term 5 ($k=4$): Coefficient is 15.

  $$15 \cdot (3)^{6-4} \cdot (5x)^4 = 15 \cdot 3^2 \cdot (625x^4) = 15 \cdot 9 \cdot 625x^4 = 15 \cdot 5625x^4 = 84375x^4$$

• Term 6 ($k=5$): Coefficient is 6.

  $$6 \cdot (3)^{6-5} \cdot (5x)^5 = 6 \cdot 3^1 \cdot (3125x^5) = 6 \cdot 3 \cdot 3125x^5 = 18 \cdot 3125x^5 = 56250x^5$$

• Term 7 ($k=6$): Coefficient is 1.

  $$1 \cdot (3)^{6-6} \cdot (5x)^6 = 1 \cdot 3^0 \cdot (15625x^6) = 1 \cdot 1 \cdot 15625x^6 = 15625x^6$$

$$(3+5x)^6 = 729 + 7290x + 30375x^2 + 67500x^3 + 84375x^4 + 56250x^5 + 15625x^6$$

The final answer is $\boxed{729 + 7290x + 30375x^2 + 67500x^3 + 84375x^4 + 56250x^5 + 15625x^6}$.