Simplify completely: $$\sqrt[3]{30^3+40^3+50^3}$$.
I recall that there is a formula for factoring the sum of two cubes: $$a^3+b^3=(a+b)(a^2-ab+b^2)$$. For example, I can write: $$30^3+40^3+50^3=(30^3+40^3)+50^3$$. Then, using the formula, I can factor the first pair: $$(30^3+40^3)+50^3=(30+40)(30^2-30\cdot 40+40^2)+50^3$$. Simplifying, I get: $$(30+40)(30^2-30\cdot 40+40^2)+50^3=70(900-1200+1600)+50^3$$. This simplifies to: $$70(1300)+50^3=91000+125000$$. Adding, I get: $$91000+125000=216000$$. So, I have simplified the expression inside the cube root to: $$\sqrt[3]{30^3+40^3+50^3}=\sqrt[ 3]{216000}$$. I look for perfect cube factors of 216000. I notice that 216 is a perfect cube, since $6^3=216$. I also notice that 1000 is a perfect cube, since $10^3=1000$. So, I can write: $$\sqrt[3]{216000}=\sqrt[3]{216\cdot 1000}$$. Using the property that the cube root of a product is the product of the cube roots, I get: $$\sqrt[3]{216\cdot 1000}=\sqrt[3]{216}\cdot \sqrt[3]{1000}$$. Taking the cube roots of 216 and 1000, I get: $$\sqrt[3]{216}\cdot \sqrt[3]{1000}=6\cdot 10$$. Multiplying, I get: $$6\cdot 10=60$$. 