What is (sqrt(16^3)) 1/4 ?

The solution to (sqrt(16^3)) 1/4 is 16

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simplify (sqrt(16^3)) 1/4

simplify

(1 6^{3}) \frac{1}{4}

16

Solution steps

(1 6^{3}) \frac{1}{4}

Apply rule:

(a) = a

(1 6^{3}) = 1 6^{3}

= 1 6^{3} \frac{1}{4}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

1 6^{2 + 1} = 1 6^{2} \cdot 16

= 1 6^{2} \cdot 16 \frac{1}{4}

Apply radical rule:

ab = a b, a \geq 0, b \geq 0

1 6^{2} \cdot 16 = 1 6^{2} 16

= 1 6^{2} 16 \frac{1}{4}

Apply radical rule:

a^{2} = a, a \geq 0

1 6^{2} = 16

= 1616 \frac{1}{4}

Prime factorization of

16 : 2^{4}

= 16 2^{4} \frac{1}{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}}

= 16 \cdot 2^{\frac{4}{2}} \cdot \frac{1}{4}

Divide the numbers:

\frac{4}{2} = 2

= 16 \cdot 2^{2} \cdot \frac{1}{4}

2^{2} = 4

= 16 \cdot 4 \cdot \frac{1}{4}

Multiply the numbers:

16 \cdot 4 = 64

= 64 \cdot \frac{1}{4}

64 \cdot \frac{1}{4} = 16

= 16

(x^{2} - 2 x - 2) (3 x^{2} - 4 x + 4)

\frac{x ^{12}}{x ^{3}}

\frac{- a ^{2} b}{- ab}

\frac{\frac{2 x + 5}{4}}{\frac{2 x + 5}{6}}

- 5 + (- 10)