Solve {left({C}_{6}right)}^0-{left({C}_{6}right)}^1+{left({C}_{6}right)}^2-{left({C}_{6}right)}^3

$${ \left( { C }_{ 6 } \right) }^{ 0 } - { \left( { C }_{ 6 } \right) }^{ 1 } + { \left( { C }_{ 6 } \right) }^{ 2 } - { \left( { C }_{ 6 } \right) }^{ 3 }$$

Evaluate

$$\left(1-C_{6}\right)\left(C_{6}^{2}+1\right)$$

Solution Steps

Calculate C_{6} to the power of 0 and get 1.

$$1-C_{6}^{1}+C_{6}^{2}-C_{6}^{3}$$

Calculate C_{6} to the power of 1 and get C_{6}.

$$1-C_{6}+C_{6}^{2}-C_{6}^{3}$$

Factor

$$\left(1-C_{6}\right)\left(C_{6}^{2}+1\right)$$

Solution Steps

Multiply and combine like terms.

$$-C_{6}^{3}+C_{6}^{2}-C_{6}+1$$

Do the grouping -C_{6}^{3}+C_{6}^{2}-C_{6}+1=\left(-C_{6}^{3}+C_{6}^{2}\right)+\left(-C_{6}+1\right), and factor out C_{6}^{2} in -C_{6}^{3}+C_{6}^{2}.

$$C_{6}^{2}\left(-C_{6}+1\right)-C_{6}+1$$

Factor out common term -C_{6}+1 by using distributive property. Polynomial C_{6}^{2}+1 is not factored since it does not have any rational roots.

$$\left(-C_{6}+1\right)\left(C_{6}^{2}+1\right)$$