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

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

Evaluate

$$-270$$

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{3}\right)^{3}.

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

The square of \sqrt{3} is 3.

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

Multiply -36 and 3 to get -108.

$$-27+54\sqrt{3}-108+8\left(\sqrt{3}\right)^{3}+\left(-3-2\sqrt{3}\right)^{3}$$

Subtract 108 from -27 to get -135.

$$-135+54\sqrt{3}+8\left(\sqrt{3}\right)^{3}+\left(-3-2\sqrt{3}\right)^{3}$$

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

$$-135+54\sqrt{3}+8\left(\sqrt{3}\right)^{3}-27-54\sqrt{3}-36\left(\sqrt{3}\right)^{2}-8\left(\sqrt{3}\right)^{3}$$

The square of \sqrt{3} is 3.

$$-135+54\sqrt{3}+8\left(\sqrt{3}\right)^{3}-27-54\sqrt{3}-36\times 3-8\left(\sqrt{3}\right)^{3}$$

Multiply -36 and 3 to get -108.

$$-135+54\sqrt{3}+8\left(\sqrt{3}\right)^{3}-27-54\sqrt{3}-108-8\left(\sqrt{3}\right)^{3}$$

Subtract 108 from -27 to get -135.

$$-135+54\sqrt{3}+8\left(\sqrt{3}\right)^{3}-135-54\sqrt{3}-8\left(\sqrt{3}\right)^{3}$$

Subtract 135 from -135 to get -270.

$$-270+54\sqrt{3}+8\left(\sqrt{3}\right)^{3}-54\sqrt{3}-8\left(\sqrt{3}\right)^{3}$$

Combine 54\sqrt{3} and -54\sqrt{3} to get 0.

$$-270+8\left(\sqrt{3}\right)^{3}-8\left(\sqrt{3}\right)^{3}$$

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

$$-270$$

Factor

$$-270$$