Solve 3*sqrt{27}-2*(1+sqrt{9}+2*sqrt{27}+sqrt{36}-2sqrt{3}-sqrt{4}-18)

$$3 \cdot \sqrt { 27 } - 2 \cdot ( 1 + \sqrt { 9 } + 2 \cdot \sqrt { 27 } + \sqrt { 36 } - 2 \sqrt { 3 } - \sqrt { 4 } - 18 )$$

Evaluate

$$\sqrt{3}+20\approx 21.732050808$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$3\times 3\sqrt{3}-2\left(1+\sqrt{9}+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-18\right)$$

Multiply 3 and 3 to get 9.

$$9\sqrt{3}-2\left(1+\sqrt{9}+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-18\right)$$

Calculate the square root of 9 and get 3.

$$9\sqrt{3}-2\left(1+3+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-18\right)$$

Add 1 and 3 to get 4.

$$9\sqrt{3}-2\left(4+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-18\right)$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$9\sqrt{3}-2\left(4+2\times 3\sqrt{3}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-18\right)$$

Multiply 2 and 3 to get 6.

$$9\sqrt{3}-2\left(4+6\sqrt{3}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-18\right)$$

Calculate the square root of 36 and get 6.

$$9\sqrt{3}-2\left(4+6\sqrt{3}+6-2\sqrt{3}-\sqrt{4}-18\right)$$

Add 4 and 6 to get 10.

$$9\sqrt{3}-2\left(10+6\sqrt{3}-2\sqrt{3}-\sqrt{4}-18\right)$$

Combine 6\sqrt{3} and -2\sqrt{3} to get 4\sqrt{3}.

$$9\sqrt{3}-2\left(10+4\sqrt{3}-\sqrt{4}-18\right)$$

Calculate the square root of 4 and get 2.

$$9\sqrt{3}-2\left(10+4\sqrt{3}-2-18\right)$$

Subtract 2 from 10 to get 8.

$$9\sqrt{3}-2\left(8+4\sqrt{3}-18\right)$$

Subtract 18 from 8 to get -10.

$$9\sqrt{3}-2\left(-10+4\sqrt{3}\right)$$

Use the distributive property to multiply -2 by -10+4\sqrt{3}.

$$9\sqrt{3}+20-8\sqrt{3}$$

Combine 9\sqrt{3} and -8\sqrt{3} to get \sqrt{3}.

$$\sqrt{3}+20$$