Solve (frac{-1+sqrt{3}}{2})^3+(frac{-1-1sqrt{3}}{2})^3=2

$$( \frac { - 1 + \sqrt { 3 } } { 2 } ) ^ { 3 } + ( \frac { - 1 - 1 \sqrt { 3 } } { 2 } ) ^ { 3 } = 2$$

Verify

Solution Steps

To raise \frac{-1+\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To raise \frac{-1-\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Since \frac{\left(-1+\sqrt{3}\right)^{3}}{2^{3}} and \frac{\left(-1-\sqrt{3}\right)^{3}}{2^{3}} have the same denominator, add them by adding their numerators.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -3 and 3 to get -9.

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

Subtract 9 from -1 to get -10.

$$\frac{-10+3\sqrt{3}+\left(\sqrt{3}\right)^{3}+\left(-1-\sqrt{3}\right)^{3}}{2^{3}}=2$$

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

$$\frac{-10+3\sqrt{3}+\left(\sqrt{3}\right)^{3}-1-3\sqrt{3}-3\left(\sqrt{3}\right)^{2}-\left(\sqrt{3}\right)^{3}}{2^{3}}=2$$

The square of \sqrt{3} is 3.

$$\frac{-10+3\sqrt{3}+\left(\sqrt{3}\right)^{3}-1-3\sqrt{3}-3\times 3-\left(\sqrt{3}\right)^{3}}{2^{3}}=2$$

Multiply -3 and 3 to get -9.

$$\frac{-10+3\sqrt{3}+\left(\sqrt{3}\right)^{3}-1-3\sqrt{3}-9-\left(\sqrt{3}\right)^{3}}{2^{3}}=2$$

Subtract 9 from -1 to get -10.

$$\frac{-10+3\sqrt{3}+\left(\sqrt{3}\right)^{3}-10-3\sqrt{3}-\left(\sqrt{3}\right)^{3}}{2^{3}}=2$$

Subtract 10 from -10 to get -20.

$$\frac{-20+3\sqrt{3}+\left(\sqrt{3}\right)^{3}-3\sqrt{3}-\left(\sqrt{3}\right)^{3}}{2^{3}}=2$$

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

$$\frac{-20+\left(\sqrt{3}\right)^{3}-\left(\sqrt{3}\right)^{3}}{2^{3}}=2$$

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

$$\frac{-20}{2^{3}}=2$$

Calculate 2 to the power of 3 and get 8.

$$\frac{-20}{8}=2$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

$$-\frac{5}{2}=2$$

Convert 2 to fraction \frac{4}{2}.

$$-\frac{5}{2}=\frac{4}{2}$$

Compare -\frac{5}{2} and \frac{4}{2}.

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