`cos^3 2theta+3cos2theta=4(cos^6theta-sin^6 theta)`

Explanation

`4(cos^6 x – sin^6 x) `
Now, upon expansion we get, `4(cos^6 x – sin^6 x) `= `4[(cos^2 x)^3 – (sin^2 x)^3] `= `4(cos^2 x – sin^2 x) (cos^4 x + sin^4 x + cos^2 x sin^2 x) `
On using the formula, `a^3 – b^3 = (a - b) (a^2 + b^2 + ab)`

= `4 cos 2x (cos^4 x + sin^4 x + cos^2 x sin^2 x + cos^2 x sin2 x – cos2 x sinx`
As we know, `cos 2x = cos^2 x – sin^2 x `


Therefore, = `4 cos 2x (cos^4 x + sin^4 x + 2 cos^2 x sin^2 x – cos^2 x sin^2 x) `

=` 4 cos 2x [(cos^2 x)^2 + (sin^2 x)^2 + 2 cos^2 x sin^2 x – cos^2 x sin^2 x]`


As we know, `a^2 + b^2 + 2ab = (a + b)^2`


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