Question

1. \sqrt3{-54x^6y^{11}}

Explanation:

Step1: Factor the radicand

First, we factor \(-54\), \(x^6\), and \(y^{11}\) into parts that are perfect cubes.

• For the coefficient: \(-54 = -27\times2 = (-3)^3\times2\)
• For the \(x\)-term: \(x^6=(x^2)^3\) (since \((x^2)^3 = x^{2\times3}=x^6\))
• For the \(y\)-term: \(y^{11}=y^{9 + 2}=(y^3)^3\times y^2\) (since \((y^3)^3=y^{3\times3}=y^9\) and \(y^9\times y^2 = y^{11}\))

So, \(\sqrt[3]{-54x^6y^{11}}=\sqrt[3]{(-3)^3\times2\times(x^2)^3\times(y^3)^3\times y^2}\)

Step2: Use the property of cube roots \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\)

We can separate the perfect cube factors from the non - perfect cube factors:
\(\sqrt[3]{(-3)^3\times2\times(x^2)^3\times(y^3)^3\times y^2}=\sqrt[3]{(-3)^3}\times\sqrt[3]{(x^2)^3}\times\sqrt[3]{(y^3)^3}\times\sqrt[3]{2y^2}\)

Step3: Simplify each cube root

• \(\sqrt[3]{(-3)^3}=-3\) (because the cube root of a perfect cube \(a^3\) is \(a\), and here \(a = - 3\))
• \(\sqrt[3]{(x^2)^3}=x^2\) (by the property \(\sqrt[3]{a^3}=a\))
• \(\sqrt[3]{(y^3)^3}=y^3\) (by the same property \(\sqrt[3]{a^3}=a\))
• The remaining factor is \(\sqrt[3]{2y^2}\)

Now, multiply the simplified perfect cube roots together:
\(-3\times x^2\times y^3\times\sqrt[3]{2y^2}=-3x^2y^3\sqrt[3]{2y^2}\)

Answer:

\(-3x^{2}y^{3}\sqrt[3]{2y^{2}}\)