Square root of $8^3$

We can reduce $8^3$ to its prime factors: $$8^3=512\implies 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$$ Because this is a square root, we are looking for groups of $2$. Similarly, if this was a cube root we would look for groups of 3: $$\boxed{2\cdot2}\boxed{2\cdot2}\boxed{2\cdot2}\boxed{2\cdot2}2$$ We have 4 groups of 2 which we will take out of the radical: $$2\cdot2\cdot2\cdot2\sqrt{2}$$ We can now simplify this as: $$\boxed{16\sqrt{2}}$$


Using the standard rules of algebra, we compute:

$$\sqrt{8^{3}} = \sqrt{8^2 \cdot8} = \sqrt{8^2}\cdot\sqrt{8} = 8\cdot\sqrt{4\cdot2} = 8\cdot\sqrt{2^2}\cdot\sqrt{2} = 16\sqrt{2}$$


$\sqrt[]{8^3}$ $= \sqrt[]{8^2\cdot8}$ $= 8\sqrt[]{8}$ $=8\sqrt[]{2\cdot4}$ $=2\cdot8\sqrt[]{2}$ $=16\sqrt[]{2}$

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Algebra Precalculus

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