need help simplifying this radical $\sqrt[35]{128y^{42}}$

Question

I am trying to figure out how to get to the solution below but have having difficulty. Can someone explain how to get to the solution.

$$\sqrt[35]{128y^{42}}$$

This is the answer but I can't figure out how to get it.

$$2^{\frac15}y^{\frac65}$$

Answer 1

Note that $((2y^6)^7)^{1/35}=(2y^6)^{1/5}$.

Now can you figure it out?

Answer 2

$$\sqrt[35]{128y^{42}}$$ $$=(128y^{42})^{\frac{1}{35}}$$ $$=(2^7 y^{7 \times6})^{\frac{1}{5 \times 7}}$$ $$=2^{\frac{7}{7 \times5}} y^{\frac{7 \times6}{7 \times5}}$$ $$=2^{\frac{1}{5}}y^{\frac{6}{5}}$$

Answer 3

$$\sqrt[35]{128 y^{42}} \iff {(128 y^{42})}^{\frac{1}{35}}$$ $$\iff {(128)}^{\frac{1}{35}} \times y^{\frac{42}{35}}$$ This simplifies to$${(2)}^{\frac{7}{35}} \times y^{\frac{6}{5}}$$ Finally, you get $$2^{\frac{1}{5}} y^{\frac{6}{5}}$$ as the answer. $$ \therefore \sqrt[35]{128 y^{42}} = 2^{\frac{1}{5}} y^{\frac{6}{5}}$$