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Three steps: (1) rewrite the LHS in terms of cos(x); (2) pull out [cos(x)]^2 as a factor; (3) show that the other factor equals 4 cos(2x).
cos(2x) = 2 [cos(x)]^2 - 1.
cos(4x)
= 2 [cos(2x)]^2 - 1
= 2 (2 [cos(x)]^2 - 1)^2 - 1
= 8 [cos(x)]^4 - 8 [cos(x)]^2 + 1.
So
1 + 2 cos(2x) + cos(4x)
= 8 [cos(x)]^4 - 4 [cos(x)]^2
= 4 [cos(x)]^2 * (2 [cos(x)]^2 - 1)
= 4 [cos(x)]^2 * cos(2x).
cos(2x) = 2 [cos(x)]^2 - 1.
cos(4x)
= 2 [cos(2x)]^2 - 1
= 2 (2 [cos(x)]^2 - 1)^2 - 1
= 8 [cos(x)]^4 - 8 [cos(x)]^2 + 1.
So
1 + 2 cos(2x) + cos(4x)
= 8 [cos(x)]^4 - 4 [cos(x)]^2
= 4 [cos(x)]^2 * (2 [cos(x)]^2 - 1)
= 4 [cos(x)]^2 * cos(2x).
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