Simplify nth Roots $6\sqrt[3]{9000} + 7\sqrt[3]{576}$

Question

I'm having some concerns about how I am going about simplifying this radical expression. I wanted to know if this would be an accurate method of solving.

Simplify: $$6\sqrt[3]{9000} + 7\sqrt[3]{576}$$

Radical expression product rule & greatest common factor: $$6\sqrt[3]{125}\sqrt[3]{72} + 7\sqrt[3]{8}\sqrt[3]{72}$$

Principle of nth root: $$6 \cdot 5\sqrt[3]{72} + 7 \cdot 2\sqrt[3]{72}$$

Simplify: $$30\sqrt[3]{72} + 14\sqrt[3]{72}$$

Addition of radical rule: $$30 + 14\sqrt[3]{72}$$

Simplify: $$44\sqrt[3]{72}$$

Is $44\sqrt[3]{72}$ completely simplified? I'm positive that 72 has no whole number $a$ that satisfies $72 = a^{3}.$

Answer 1

You write "I'm positive that 72 has no whole number $a$ that satisfies $72 = a^{3}$". However, simplifying also works if there $72$ has a cubic factor. And indeed, $72$ contains a cube $a^3=8$. So we get $88\cdot \sqrt[3]{9}$.

Answer 2

$$ \sqrt[3]{9000} = (9 \times 10^3)^{\frac{1}{3}} = 10 \times 9^{\frac{1}{3}}$$

$$ \sqrt[3]{576} = \sqrt[3]{(2^3 \times 3)^2} = 4 \times 9^{\frac{1}{3}}$$

$$ 6\sqrt[3]{9000} = 60 \times 9^{\frac{1}{3}}$$

$$ 7\sqrt[3]{576} = 28 \times 9^{\frac{1}{3}} $$

$$ 7\sqrt[3]{576} + 6\sqrt[3]{9000} = (60 \times 9^{\frac{1}{3}}) + (28 \times 9^{\frac{1}{3}}) = 88 \times 9^{\frac{1}{3}} $$