Simplify √27 − √2(2√6 − √8).

Step-by-step solution

1. 1

   Distribute √2 across the bracket before simplifying anything else. Multiplying surds combines the radicands: $\sqrt a\cdot\sqrt b=\sqrt{ab}$. Simplifying $\sqrt6$ and $\sqrt8$ first would work too, but distributing keeps the products tidy:

   $\sqrt2\cdot 2\sqrt6=2\sqrt{12},\qquad \sqrt2\cdot\sqrt8=\sqrt{16}$

2. 2

   Simplify the two products.

   $2\sqrt{12}=2\cdot 2\sqrt3=4\sqrt3,\qquad \sqrt{16}=4$

   so the bracket contributes $\sqrt2\left(2\sqrt6-\sqrt8\right)=4\sqrt3-4$.

3. 3

   Simplify the leading term.

   $\sqrt{27}=\sqrt{9\cdot 3}=3\sqrt3$

4. 4

   Subtract, distributing the minus sign over both terms.

   $3\sqrt3-\left(4\sqrt3-4\right)=3\sqrt3-4\sqrt3+4=-\sqrt3+4$

   The sign flip on the $-4$ is the step most often missed.

5. 5

   State the answer.

   $\boxed{4-\sqrt3}$

6. 6

   Check numerically. $\sqrt{27}=5.1962$, and $\sqrt2(2\sqrt6-\sqrt8)=1.4142(4.8990-2.8284)=1.4142(2.0706)=2.9282$. Then $5.1962-2.9282=2.2680$, and $4-\sqrt3=4-1.7321=2.2679$ ✓.