Answer:
happy 2021 this willl be a good year
<span>cos x = [ e^(ix) - e^(-ix) ] / 2i along with [ A - B ]^5 =
( A^5 - B^5) - 5 AB ( A^3 - B^3) + 10 A²B² (A - B )...A = e^(ix) , B = e^(-ix)
[ cos x ] ^5 = (1/32i) { [ e^(5ix) - e^(-5x) ] - 5 [ ( e^(3ix) - e^(-3ix) ] + 10 [ e^(ix) - e^(-ix) ] }
= ( 1 / 16 ) { cos 5x - 5 cos 3x + 10 cos x }.....much faster</span>
When x = 0 y = 0 so the graph has an intercept at the origin ( 0, 0)
let x^5 + 10x - 9x = 0
x^5 + x = 0
x(x^4 + 1) = 0
x^4 = - 1 so there are no other real roots
There is only 1 intercept at (0, 0) answer
