Solve frac{{left(sqrt{6}+sqrt{3}*1right)}^2}{3^2*-1}

$$\frac{ { \left( \sqrt{ 6 } + \sqrt{ 3 } \times 1 \right) }^{ 2 } }{ { 3 }^{ 2 } \times -1 }$$

Evaluate

$$-\frac{2\sqrt{2}}{3}-1\approx -1.942809042$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+1\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

The square of \sqrt{6} is 6.

$$\frac{6+2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{6+2\sqrt{3}\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{6+2\times 3\sqrt{2}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

Multiply 2 and 3 to get 6.

$$\frac{6+6\sqrt{2}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

The square of \sqrt{3} is 3.

$$\frac{6+6\sqrt{2}+3}{3^{2}\left(-1\right)}$$

Add 6 and 3 to get 9.

$$\frac{9+6\sqrt{2}}{3^{2}\left(-1\right)}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{9+6\sqrt{2}}{9\left(-1\right)}$$

Multiply 9 and -1 to get -9.

$$\frac{9+6\sqrt{2}}{-9}$$

Expand

$$-\frac{2 \sqrt{2}}{3} - 1 = -1.942809042$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+1\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

The square of \sqrt{6} is 6.

$$\frac{6+2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{6+2\sqrt{3}\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{6+2\times 3\sqrt{2}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

Multiply 2 and 3 to get 6.

$$\frac{6+6\sqrt{2}+\left(\sqrt{3}\right)^{2}}{3^{2}\left(-1\right)}$$

The square of \sqrt{3} is 3.

$$\frac{6+6\sqrt{2}+3}{3^{2}\left(-1\right)}$$

Add 6 and 3 to get 9.

$$\frac{9+6\sqrt{2}}{3^{2}\left(-1\right)}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{9+6\sqrt{2}}{9\left(-1\right)}$$

Multiply 9 and -1 to get -9.

$$\frac{9+6\sqrt{2}}{-9}$$