Prove that $6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0​$ Prove that
​$$6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0​$$
Expression can be verified for different values of $A$ such as $\frac\pi4,\frac\pi2$ etc. But to prove it for general value of A?
 A: Hint: Try to use $\cos^2 A = 1-\sin^2 A $

 Writing $s$ for $\sin^2 A$ we have $6(s^5 + (1-s)^5) -15(s^4+ (1-s)^4) + 10(s^3 + (1-s)^3) - 1 = 6s^5 + 6 - 30s + 60s^2 - 60s^3 + 30s^4 - 6s^5 - 15s^4 - 15 + 60s - 90s^2 + 60s^3 - 15s^4 + 10s^3 + 10 - 30s + 30s^2 - 10s^3 - 1 = 0$

A: Set $x=\sin^2A$ and $y=\cos^2A$. Then $x+y=1$ and
\begin{align}
1&=(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\\
&=x^5+y^5+5xy(x^3+y^3)+10x^2y^2(x+y)\\
&=x^5+y^5+5xy(x+y)(x^2-xy+y^2)+10x^2y^2\\
&=x^5+y^5+5xy(x^2-xy+y^2)+10x^2y^2\\
&=x^5+y^5+5xy((x+y)^2-3xy)+10x^2y^2\\
&=x^5+y^5+5xy(1-3xy)+10x^2y^2\\
&=x^5+y^5-5x^2y^2+5xy
\end{align}
Similarly,
\begin{align}
1&=(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\
&=x^4+y^4+4xy(x^2+y^2)+6x^2y^2\\
&=x^4+y^4+4xy(1-2xy)+6x^2y^2\\
&=x^4+y^4-2x^2y^2+4xy
\end{align}
and
\begin{align}
1&=(x+y)^3=x^3+3x^2y+3xy^2+y^3\\
&=x^3+y^3+3xy(x+y)\\
&=x^3+y^3+3xy
\end{align}
Set $p=xy$; we have proved that
\begin{align}
x^5+y^5&=5p^2-5p+1\\
x^4+y^4&=2p^2-4p+1\\
x^3+y^3&=-3p+1
\end{align}
so finally your expression is
$$
6(5p^2-5p+1)-15(2p^2-4p+1)+10(-3p+1)-1=0
$$
