Solve (3+2sqrt{2})*(3-2sqrt{2})^2

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

Evaluate

$$3-2\sqrt{2}\approx 0.171572875$$

Solution Steps

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

$$\left(3+2\sqrt{2}\right)\left(9-12\sqrt{2}+4\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$\left(3+2\sqrt{2}\right)\left(9-12\sqrt{2}+4\times 2\right)$$

Multiply 4 and 2 to get 8.

$$\left(3+2\sqrt{2}\right)\left(9-12\sqrt{2}+8\right)$$

Add 9 and 8 to get 17.

$$\left(3+2\sqrt{2}\right)\left(17-12\sqrt{2}\right)$$

Use the distributive property to multiply 3+2\sqrt{2} by 17-12\sqrt{2} and combine like terms.

$$51-2\sqrt{2}-24\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$51-2\sqrt{2}-24\times 2$$

Multiply -24 and 2 to get -48.

$$51-2\sqrt{2}-48$$

Subtract 48 from 51 to get 3.

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