1. **State the problem:** We are given $N = 2 \times 3^4 \times y^3$ and need to find $3N^2$ expressed as a product of prime factors in index form, in terms of $y$. 2. **Recall the formula and rules:** - When squaring a product, square each factor: $\left(a \times b \times c\right)^2 = a^2 \times b^2 \times c^2$. - When multiplying powers with the same base, add exponents: $a^m \times a^n = a^{m+n}$. 3. **Calculate $N^2$:** $$N^2 = \left(2 \times 3^4 \times y^3\right)^2 = 2^2 \times \left(3^4\right)^2 \times \left(y^3\right)^2$$ 4. **Simplify each term:** - $2^2 = 4$ - $\left(3^4\right)^2 = 3^{4 \times 2} = 3^8$ - $\left(y^3\right)^2 = y^{3 \times 2} = y^6$ So, $$N^2 = 4 \times 3^8 \times y^6$$ 5. **Multiply by 3:** $$3N^2 = 3 \times 4 \times 3^8 \times y^6 = 4 \times 3^{1+8} \times y^6 = 4 \times 3^9 \times y^6$$ 6. **Express 4 as prime factors:** $$4 = 2^2$$ 7. **Final expression:** $$3N^2 = 2^2 \times 3^9 \times y^6$$ **Answer:** $3N^2 = 2^2 \times 3^9 \times y^6$