$\frac{14}{\sqrt[3]{25 x^8}}$
Answer (1)
Rewrite the expression with fractional exponents.
Simplify the denominator by expressing 25 as 5 2 and distributing the exponent.
Rationalize the denominator by multiplying both numerator and denominator by a suitable term.
Simplify the expression to obtain the final answer: 5 x 3 14 3 5 x .
Explanation
Understanding the Problem We are given the expression 3 25 x 8 14 and our goal is to simplify it.
Rewriting with Fractional Exponents First, we rewrite the cube root in the denominator using fractional exponents: 3 25 x 8 14 = ( 25 x 8 ) 3 1 14
Expressing 25 as a Power of 5 Next, we express 25 as 5 2 : ( 25 x 8 ) 3 1 14 = ( 5 2 x 8 ) 3 1 14
Distributing the Exponent Now, we distribute the exponent 3 1 to both 5 2 and x 8 : ( 5 2 x 8 ) 3 1 14 = 5 3 2 x 3 8 14
Separating the Exponent of x We can rewrite x 3 8 as x 2 + 3 2 = x 2 x 3 2 : 5 3 2 x 3 8 14 = 5 3 2 x 2 x 3 2 14 = x 2 ( 5 2 x 2 ) 3 1 14 = x 2 ( 5 x 3 4 ) 3 2 14
Rationalizing the Denominator To rationalize the denominator, we want to eliminate the fractional exponent. We have x 3 8 in the denominator. We can rewrite the expression as: 5 3 2 x 3 8 14 = x 2 ( 5 2 ) 3 1 x 3 2 14 = x 2 ( 25 x 2 ) 3 1 14 To rationalize, we multiply the numerator and denominator by ( 5 x ) 3 1 :
x 2 ( 25 x 2 ) 3 1 14 = x 2 ( 5 2 x 2 ) 3 1 14 × ( 5 x ) 3 1 ( 5 x ) 3 1 = x 2 ( 5 2 x 2 ) 3 1 ( 5 x ) 3 1 14 ( 5 x ) 3 1 = x 2 ( 5 3 x 3 ) 3 1 14 ( 5 x ) 3 1 = x 2 ( 5 x ) 14 ( 5 x ) 3 1 = 5 x 3 14 ( 5 x ) 3 1
Rewriting as a Cube Root Finally, we rewrite the fractional exponent as a cube root: 5 x 3 14 ( 5 x ) 3 1 = 5 x 3 14 3 5 x
Final Answer Thus, the simplified expression is 5 x 3 14 3 5 x .
Examples
Imagine you are calculating the flow rate of a liquid through a pipe, and the formula involves simplifying an expression with radicals and exponents. Simplifying such expressions, as we did here, allows for easier computation and a clearer understanding of how different variables affect the flow rate. This is also applicable in electrical engineering when dealing with impedance calculations or in acoustics when analyzing sound wave propagation.
