# Question #b7e2c

Feb 5, 2016

You would like to prove that

$2 \left({\sin}^{6} \Theta + {\cos}^{6} \Theta\right) - 3 \left({\sin}^{4} \Theta + {\cos}^{4} \Theta\right) + 1 = 0$

This can be proven by starting at the left side and transforming until we have the right side, so in this case, $0$.

I will use the following formulas and identites:

[1] $\text{ } {\sin}^{2} \Theta + {\cos}^{2} \Theta = 1$
[2]$\text{ } {\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$
[3]$\text{ } {\left(a - b\right)}^{3} = {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3}$

$2 \left({\sin}^{6} \Theta + {\cos}^{6} \Theta\right) - 3 \left({\sin}^{4} \Theta + {\cos}^{4} \Theta\right) + 1$

$= 2 \left({\left[{\sin}^{2} \Theta\right]}^{3} + {\cos}^{6} \Theta\right) - 3 \left({\left[{\sin}^{2} \Theta\right]}^{2} + {\cos}^{4} \Theta\right) + 1$

... use [1 ]$\text{ "<=> " } {\sin}^{2} \Theta = 1 - {\cos}^{2} \Theta$ ...

$= 2 \left({\left[1 - {\cos}^{2} \Theta\right]}^{3} + {\cos}^{6} \Theta\right) - 3 \left({\left[1 - {\cos}^{2} \Theta\right]}^{2} + {\cos}^{4} \Theta\right) + 1$

... use the binomial formulas [2] and [3] ....

$= 2 \left(\left[1 - 3 {\cos}^{2} \Theta + 3 {\cos}^{4} \Theta - {\cos}^{6} \Theta\right] + {\cos}^{6} \Theta\right) - 3 \left(\left[1 - 2 {\cos}^{2} \Theta + {\cos}^{4} \Theta\right] + {\cos}^{4} \Theta\right) + 1$

$= 2 \left(1 - 3 {\cos}^{2} \Theta + 3 {\cos}^{4} \Theta \cancel{- {\cos}^{6} \Theta + {\cos}^{6} \Theta}\right) - 3 \left(1 - 2 {\cos}^{2} \Theta + 2 {\cos}^{4} \Theta\right) + 1$

$= 2 - 6 {\cos}^{2} \Theta + 6 {\cos}^{4} \Theta - 3 + 6 {\cos}^{2} \Theta - 6 {\cos}^{4} \Theta + 1$

$= 2 \cancel{\textcolor{g r e e n}{- 6 {\cos}^{2} \Theta}} \cancel{\textcolor{b l u e}{+ 6 {\cos}^{4} \Theta}} - 3 \cancel{\textcolor{g r e e n}{+ 6 {\cos}^{2} \Theta}} \cancel{\textcolor{b l u e}{- 6 {\cos}^{4} \Theta}} + 1$

$= 2 - 3 + 1$

$= 0$

q.e.d.