Solve sqrt{27}-2*(1+sqrt{9}+2*sqrt{27}+sqrt{36}-2*sqrt{3}-sqrt{4}-sqrt{64})

$$\sqrt { 27 } - 2 \cdot ( 1 + \sqrt { 9 } + 2 \cdot \sqrt { 27 } + \sqrt { 36 } - 2 \cdot \sqrt { 3 } - \sqrt { 4 } - \sqrt { 64 } )$$

Evaluate

$$-5\sqrt{3}\approx -8.660254038$$

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}.

$$3\sqrt{3}-2\left(1+\sqrt{9}+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

Calculate the square root of 9 and get 3.

$$3\sqrt{3}-2\left(1+3+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

Add 1 and 3 to get 4.

$$3\sqrt{3}-2\left(4+2\sqrt{27}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

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}.

$$3\sqrt{3}-2\left(4+2\times 3\sqrt{3}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

Multiply 2 and 3 to get 6.

$$3\sqrt{3}-2\left(4+6\sqrt{3}+\sqrt{36}-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

Calculate the square root of 36 and get 6.

$$3\sqrt{3}-2\left(4+6\sqrt{3}+6-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

Add 4 and 6 to get 10.

$$3\sqrt{3}-2\left(10+6\sqrt{3}-2\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

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

$$3\sqrt{3}-2\left(10+4\sqrt{3}-\sqrt{4}-\sqrt{64}\right)$$

Calculate the square root of 4 and get 2.

$$3\sqrt{3}-2\left(10+4\sqrt{3}-2-\sqrt{64}\right)$$

Subtract 2 from 10 to get 8.

$$3\sqrt{3}-2\left(8+4\sqrt{3}-\sqrt{64}\right)$$

Calculate the square root of 64 and get 8.

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

Subtract 8 from 8 to get 0.

$$3\sqrt{3}-2\times 4\sqrt{3}$$

Multiply 2 and 4 to get 8.

$$3\sqrt{3}-8\sqrt{3}$$

Combine 3\sqrt{3} and -8\sqrt{3} to get -5\sqrt{3}.

$$-5\sqrt{3}$$