Question

simplify. multiply and remove all perfect squares from inside the square roots. assume y is positive. \\(sqrt{12} cdot sqrt{y^3} cdot sqrt{6y} = \\)

Explanation:

Step1: Use property of square roots $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$

$\sqrt{12}\cdot\sqrt{y^{3}}\cdot\sqrt{6y}=\sqrt{12\times y^{3}\times 6y}$

Step2: Simplify the product inside the square root

First, multiply the coefficients: $12\times6 = 72$. Then, multiply the variables: $y^{3}\times y=y^{4}$. So we get $\sqrt{72y^{4}}$

Step3: Factor out perfect squares from the square root

We know that $72 = 36\times2$ and $y^{4}=(y^{2})^{2}$. So $\sqrt{72y^{4}}=\sqrt{36\times2\times(y^{2})^{2}}$

Step4: Use property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ again

$\sqrt{36\times2\times(y^{2})^{2}}=\sqrt{36}\cdot\sqrt{(y^{2})^{2}}\cdot\sqrt{2}$

Step5: Simplify the square roots of perfect squares

Since $\sqrt{36} = 6$ and $\sqrt{(y^{2})^{2}}=y^{2}$ (because $y$ is positive), we have $6\times y^{2}\times\sqrt{2}=6y^{2}\sqrt{2}$

Answer:

$6y^{2}\sqrt{2}$