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Chapter 0: Problem 68

Simplify the radical expressions if possible. $$\sqrt[3]{150}$$

Short Answer

Expert verified
\(\sqrt[3]{150}\)

Step by step solution

01

Prime Factorization of 150

Express 150 as a product of its prime factors. It can be done by dividing 150 by prime numbers (2, 3, 5, 7, 11, ...) starting from 2 up to the point where no further division is possible. Doing this provides the prime factors of 150 as \(2 \times 3 \times 5^2\).
02

Grouping the Factors

According to the properties of radicals, the cubic root can simplify to an expression \(x\) if we can group the factors into three similar terms under the root sign. Here in \(2 \times 3 \times 5^2\), we can't find any such group of three similar factors, so the expression can't be simplified further.
03

Write down the Simplified Form

Since there are no groups of three similar factors, the simplified form remains as the original, which is \(\sqrt[3]{150}\).

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Most popular questions from this chapter

a. A mathematics professor recently purchased a birthday cake for her son with the inscription $$\text { Happy }\left(2^{\frac{5}{2}} \cdot 2^{\frac{3}{4}} \div 2^{\frac{1}{4}}\right) \text { th Birthday. }$$ How old is the son? b. The birthday boy, excited by the inscription on the cake, tried to wolf down the whole thing Professor Mom, concerned about the possible metamorphosis of her son into a blimp, exclaimed, "Hold on! It is your birthday, so why not take \(\frac{8^{-\frac{4}{3}}+2^{-2}}{16^{-\frac{3}{4}}+2^{-1}}\) of the cake? I'll eat half of what's left over." How much of the cake did the professor eat?

Factor completely. $$ x^{4}-y^{4}-2 x^{3} y+2 x y^{3} $$

Simplify each expression. Assume that all variables represent positive numbers. $$ \left(8 x^{-6} y^{5}\right)^{\frac{1}{3}}\left(x^{\frac{5}{6}} y^{-\frac{1}{3}}\right)^{6} $$

Simplify using properties of exponents. $$\frac{\left(2 y^{\frac{1}{5}}\right)^{4}}{y^{\frac{3}{10}}}$$

Factor completely. $$ (x-5)^{-\frac{1}{2}}(x+5)^{-\frac{1}{2}}-(x+5)^{\frac{1}{2}}(x-5)^{-\frac{3}{2}} $$
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