Question

19
simplify: $sqrt3{6x^2} cdot sqrt3{6x^2}$
a $xsqrt3{36x}$
b $6xsqrt3{x}$
c $6x^2$
d $xsqrt3{12x}$

Explanation:

Step1: Recall the property of cube roots

For any real numbers \(a\) and \(b\), \(\sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab}\). So we can apply this property to \(\sqrt[3]{6x^{2}}\cdot\sqrt[3]{6x^{2}}\).

$$ \sqrt[3]{6x^{2}}\cdot\sqrt[3]{6x^{2}}=\sqrt[3]{(6x^{2})\cdot(6x^{2})} $$

Step2: Multiply the terms inside the cube root

Multiply \(6\) and \(6\), and \(x^{2}\) and \(x^{2}\) using the rule \(a^{m}\cdot a^{n}=a^{m + n}\).

$$ (6x^{2})\cdot(6x^{2})=6\times6\times x^{2+2}=36x^{4} $$

So now we have \(\sqrt[3]{36x^{4}}\).

Step3: Simplify the cube root

We can rewrite \(x^{4}\) as \(x^{3}\cdot x\) (since \(x^{3}\cdot x=x^{3 + 1}=x^{4}\)). Then:

$$ \sqrt[3]{36x^{4}}=\sqrt[3]{36x^{3}\cdot x} $$

Using the property \(\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}\) again, we get:

$$ \sqrt[3]{36x^{3}\cdot x}=\sqrt[3]{x^{3}}\cdot\sqrt[3]{36x} $$

Since \(\sqrt[3]{x^{3}} = x\) (for real \(x\)), this simplifies to:

$$ x\sqrt[3]{36x} $$

Answer:

A. \(x\sqrt[3]{36x}\)