Solve (sqrt{27}+sqrt{3}-1)^2

$$( \sqrt { 27 } + \sqrt { 3 } - 1 ) ^ { 2 }$$

Evaluate

$$49-8\sqrt{3}\approx 35.143593539$$

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}.

$$\left(3\sqrt{3}+\sqrt{3}-1\right)^{2}$$

Combine 3\sqrt{3} and \sqrt{3} to get 4\sqrt{3}.

$$\left(4\sqrt{3}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{3}-1\right)^{2}.

$$16\left(\sqrt{3}\right)^{2}-8\sqrt{3}+1$$

The square of \sqrt{3} is 3.

$$16\times 3-8\sqrt{3}+1$$

Multiply 16 and 3 to get 48.

$$48-8\sqrt{3}+1$$

Add 48 and 1 to get 49.

$$49-8\sqrt{3}$$

Expand

$$49-8\sqrt{3}$$

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}.

$$\left(3\sqrt{3}+\sqrt{3}-1\right)^{2}$$

Combine 3\sqrt{3} and \sqrt{3} to get 4\sqrt{3}.

$$\left(4\sqrt{3}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{3}-1\right)^{2}.

$$16\left(\sqrt{3}\right)^{2}-8\sqrt{3}+1$$

The square of \sqrt{3} is 3.

$$16\times 3-8\sqrt{3}+1$$

Multiply 16 and 3 to get 48.

$$48-8\sqrt{3}+1$$

Add 48 and 1 to get 49.

$$49-8\sqrt{3}$$