Prove the following: cos6x = 32cos⁶x - 48cos⁴x + 18cos²x

Use cos3x = 4cos³x - 3cosx
LHS = cos6x = cos3(2x) = 4cos³(2x) - 3cos(2x) = 4(2cos²x - 1)³ - 3(2cos²x - 1)
= 4(8cos⁶x - 12cos⁴x + 6cos²x - 1) - 6cos²x + 3
= 32cos⁶x - 48cos⁴x + 24cos²x - 4 - 6cos²x + 3
= 32cos⁶x - 48cos⁴x + 18cos²x - 1
There seems to be a mistake in the question. The correct identity is:
cos(6x) = 32cos⁶x - 48cos⁴x + 18cos²x - 1
Let's verify this using the triple angle formula:
cos(3x) = 4cos³x - 3cosx
Let x = 2x:
cos(6x) = 4cos³(2x) - 3cos(2x)
= 4(2cos²x - 1)³ - 3(2cos²x - 1)
= 4(8cos⁶x - 12cos⁴x + 6cos²x - 1) - 6cos²x + 3
= 32cos⁶x - 48cos⁴x + 24cos²x - 4 - 6cos²x + 3
= 32cos⁶x - 48cos⁴x + 18cos²x - 1
Therefore, cos6x = 32cos⁶x - 48cos⁴x + 18cos²x - 1