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

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

Evaluate

$$99$$

Solution Steps

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

$$\left(27+54\sqrt{2}+36\left(\sqrt{2}\right)^{2}+8\left(\sqrt{2}\right)^{3}+\left(3-2\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

The square of \sqrt{2} is 2.

$$\left(27+54\sqrt{2}+36\times 2+8\left(\sqrt{2}\right)^{3}+\left(3-2\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Multiply 36 and 2 to get 72.

$$\left(27+54\sqrt{2}+72+8\left(\sqrt{2}\right)^{3}+\left(3-2\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Add 27 and 72 to get 99.

$$\left(99+54\sqrt{2}+8\left(\sqrt{2}\right)^{3}+\left(3-2\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

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

$$\left(99+54\sqrt{2}+8\left(\sqrt{2}\right)^{3}+27-54\sqrt{2}+36\left(\sqrt{2}\right)^{2}-8\left(\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

The square of \sqrt{2} is 2.

$$\left(99+54\sqrt{2}+8\left(\sqrt{2}\right)^{3}+27-54\sqrt{2}+36\times 2-8\left(\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Multiply 36 and 2 to get 72.

$$\left(99+54\sqrt{2}+8\left(\sqrt{2}\right)^{3}+27-54\sqrt{2}+72-8\left(\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Add 27 and 72 to get 99.

$$\left(99+54\sqrt{2}+8\left(\sqrt{2}\right)^{3}+99-54\sqrt{2}-8\left(\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Add 99 and 99 to get 198.

$$\left(198+54\sqrt{2}+8\left(\sqrt{2}\right)^{3}-54\sqrt{2}-8\left(\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Combine 54\sqrt{2} and -54\sqrt{2} to get 0.

$$\left(198+8\left(\sqrt{2}\right)^{3}-8\left(\sqrt{2}\right)^{3}\right)\times \frac{1}{2}$$

Combine 8\left(\sqrt{2}\right)^{3} and -8\left(\sqrt{2}\right)^{3} to get 0.

$$198\times \frac{1}{2}$$

Multiply 198 and \frac{1}{2} to get 99.

$$99$$

Factor

$$3^{2}\times 11$$