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

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

Evaluate

$$17\left(2\sqrt{2}+3\right)\approx 99.083261121$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

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

Add 9 and 8 to get 17.

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

Use the distributive property to multiply 3 by 17+12\sqrt{2}.

$$51+36\sqrt{2}-2\sqrt{2}$$

Combine 36\sqrt{2} and -2\sqrt{2} to get 34\sqrt{2}.

$$51+34\sqrt{2}$$