2cosa = x + 1/x

We need prove that :

2cos3x = x^3 + `1/x^3

We know that:

cos3a = 4cos^3 a - 3cos a

Now multiply by 2:

2cos3a = 8cos^3 a - 6cos a

= (2cos a)^3 - 3*2cos a

But 2cos a = x + 1/x

==> 2cos3x = (x+ 1/x)^3 - 3(x+1/x)

Open brackets:

==> 2cos3a= x^3 + 3x^2(1/x) + 3x(1/x^2) + 1/x^3 - 3(x + 1/x)

x^3 + 3x + 3/x + 1/x^3 - 3x - 3/x

Now r3educe similar:

==> cos 3a = x^3 + 1/x^3

We'll write the formula for cos 3a:

cos 3a = 4(cos a)^3 - 3 cos a

Now, we'll substitute the formula into the expression:

2 cos 3a = 2*[ 4(cos a)^3 - 3 cos a] = 8(cos a)^3 - 6 cos a

We could write 8(cos a)^3 = (2*cos a)^3

But 2cos a = x + 1/x

We'll substitute 2cos a = x + 1/x into the expression:

(2*cos a)^3- 3*2 cos a = (x + 1/x)^3 - 3*(x + 1/x)

We'll expand the cube:

(x + 1/x)^3 = x^3 + (1/x)^3 + 3*x*(1/x)*(x + 1/x)- 3*(x + 1/x)

We'll elimiknate like terms:

**2 cos 3a = (x + 1/x)^3 = x^3 + (1/x)^3**

Given 2cosa = x+1/x. 2 cos3a = x^3+1/x^3

We know that cos3a = 4cos^3 a-3cosa

cos3a = 4[(x+1/x)/2]^3 - 3 [(x+1/x)/2], as cosa =

cos3a = (4/8)(x^3+3x^2/x+3x/x^2+1/x^3)-(3/2)(x+1/x)

cos3a = (1/2){{x^3 +3x+3/x+1) -3(x+1/x)}

cos3a = (1/2)(x^3+ 1/x^3), as the other terms cancel.

2cos3a = x^3+1/x^3.