Simplify. $\newline$ Multiply and remove all perfect squares from inside the square roots. Assume $b$ is positive. $\newline$ $\sqrt{24 b^{3}} \cdot \sqrt{40 b^{2}} \cdot \sqrt{b^{2}}=$ $\newline$ $\square$

Full solution

Q. Simplify.

\newline

Multiply and remove all perfect squares from inside the square roots. Assume

b

is positive.

\newline

\sqrt{24 b^{3}} \cdot \sqrt{40 b^{2}} \cdot \sqrt{b^{2}}=

\newline

\square

Multiply Square Roots: First, let's multiply the square roots together. $\sqrt{24b^{3}} \times \sqrt{40b^{2}} \times \sqrt{b^{2}} = \sqrt{24 \times 40 \times b^{3} \times b^{2} \times b^{2}}$
Combine Numbers and Exponents: Now, let's multiply the numbers and add the exponents for $b$ . $\sqrt{24 \times 40 \times b^{3+2+2}} = \sqrt{960 \times b^7}$
Simplify Square Root of $960$ : Next, we can simplify $\sqrt{960}$ by finding the perfect square factors of $960$ . $\newline$ $960 = 64 \times 15$ , and $64$ is a perfect square because $\sqrt{64} = 8$ . $\newline$ So, $\sqrt{960} = \sqrt{64 \times 15} = \sqrt{64} \times \sqrt{15} = 8 \times \sqrt{15}$
Simplify $b^7$ Under Square Root: Now, let's simplify $b^7$ under the square root. $b^7 = b^6 \cdot b^1$ , and $b^6$ is a perfect square because $\sqrt{b^6} = b^3$ . So, $\sqrt{b^7} = \sqrt{b^6 \cdot b} = \sqrt{b^6} \cdot \sqrt{b} = b^3 \cdot \sqrt{b}$
Combine Simplified Square Roots: Finally, we combine the simplified square roots. $8 \sqrt{15} b^3 \sqrt{b} = 8b^3 \sqrt{15b}$