Question

Rationalize denominator when a monomial is in the denominator.Please show steps

Answer 1

Answer:

$\frac{\sqrt[3]{90 x^2 y z^2} }{6 y z}$

Step-by-step explanation:

step 1;-

Given $\frac{\sqrt[3]{5 x^2} }{\sqrt[3]{12 y^2 z} }$

now you have rationalizing  denominator  (i.e monomial) with

$\frac{\sqrt[3]{5 x^2} }{\sqrt[3]{12 y^2 z} } X \frac{\sqrt[3]{(12 y^2 z)^{2} } }{\sqrt[3]{(12 y^2 z)^2} }$

By using algebraic formula is

• $\sqrt{ab} = \sqrt{a} \sqrt{b}$......(a)

• now $\frac{\sqrt[3]{5 x^2)(12 y^2 z)^2} }{\sqrt[3]{12 y^2 z)(12 y^2 z)^2} }$

• $\frac{\sqrt[3]{720 x^2 y^4 z^2} }{\sqrt[3]{(12 y^2 z)^{3} } }$....(1)

• again using Formula $\sqrt[n]{a^{n} } =a$

now simplification , we get denominator function

• $\frac{\sqrt[3]{720 x^2 y^4 z^2} }{12 y^2 z}$

again you have to simplify numerator term

• $\frac{\sqrt[3]{2^3 y^3 90 (x^2 y z^2)} }{12 y^2 z}$

now simplify

• $\frac{2 y\sqrt[3]{90 x^2 y z^2} }{12 y^2 z}$

        cancelling y and 2 values

we get Final answer

$\frac{\sqrt[3]{90 x^2 y z^2} }{6 y z}$