Solve {left(frac{-1+sqrt{5}}{2}right)}^3-2(frac{-1+sqrt{5}}{2})+1=0

$${ \left( \frac{ -1+ \sqrt{ 5 } }{ 2 } \right) }^{ 3 } -2( \frac{ -1+ \sqrt{ 5 } }{ 2 } )+1=0$$

Verify

Short Solution Steps

Multiply both sides of the equation by 2.

$$2\left(\left(\frac{-1+\sqrt{5}}{2}\right)^{3}-2\times \frac{-1+\sqrt{5}}{2}\right)+2=0$$

To raise \frac{-1+\sqrt{5}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$2\left(\frac{\left(-1+\sqrt{5}\right)^{3}}{2^{3}}-2\times \frac{-1+\sqrt{5}}{2}\right)+2=0$$

Cancel out 2 and 2.

$$2\left(\frac{\left(-1+\sqrt{5}\right)^{3}}{2^{3}}-\left(-1+\sqrt{5}\right)\right)+2=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -1+\sqrt{5} times \frac{2^{3}}{2^{3}}.

$$2\left(\frac{\left(-1+\sqrt{5}\right)^{3}}{2^{3}}-\frac{\left(-1+\sqrt{5}\right)\times 2^{3}}{2^{3}}\right)+2=0$$

Since \frac{\left(-1+\sqrt{5}\right)^{3}}{2^{3}} and \frac{\left(-1+\sqrt{5}\right)\times 2^{3}}{2^{3}} have the same denominator, subtract them by subtracting their numerators.

$$2\times \frac{\left(-1+\sqrt{5}\right)^{3}-\left(-1+\sqrt{5}\right)\times 2^{3}}{2^{3}}+2=0$$

Do the multiplications in \left(-1+\sqrt{5}\right)^{3}-\left(-1+\sqrt{5}\right)\times 2^{3}.

$$2\times \frac{-1+3\sqrt{5}+3\left(-1\right)\left(\sqrt{5}\right)^{2}+\left(\sqrt{5}\right)^{3}+8-8\sqrt{5}}{2^{3}}+2=0$$

Do the calculations in -1+3\sqrt{5}+3\left(-1\right)\left(\sqrt{5}\right)^{2}+\left(\sqrt{5}\right)^{3}+8-8\sqrt{5}.

$$2\times \frac{-8}{2^{3}}+2=0$$

Express 2\times \frac{-8}{2^{3}} as a single fraction.

$$\frac{2\left(-8\right)}{2^{3}}+2=0$$

Cancel out 2 in both numerator and denominator.

$$\frac{-8}{2^{2}}+2=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{2^{2}}{2^{2}}.

$$\frac{-8}{2^{2}}+\frac{2\times 2^{2}}{2^{2}}=0$$

Since \frac{-8}{2^{2}} and \frac{2\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{-8+2\times 2^{2}}{2^{2}}=0$$

Do the multiplications in -8+2\times 2^{2}.

$$\frac{-8+8}{2^{2}}=0$$

Do the calculations in -8+8.

$$\frac{0}{2^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{0}{4}=0$$

Zero divided by any non-zero number gives zero.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$