If $7 \theta = (2n + 1)\pi$, when $n = 0, 1, 2, 3, 4, 5, 6,$ then o n the basis of given information, answer the given question. The equation whose roots are $\cos\dfrac{\pi }7, \cos \dfrac{3\pi }7, \cos \dfrac{5\pi}7$ is,

Using the identity for cos(7theta) = 64cos^7(theta) - 112cos^5(theta) + 56cos^3(theta) - 7cos(theta), we can write the expansion for 7theta = (2n + 1)pi. Letting x = cos(theta) and setting the equation to zero gives 64x^7 - 112x^5 + 56x^3 - 7x + 1 = 0. This polynomial has roots corresponding to n = 0 to 6, and factoring out the known roots x = 1 (for n = 3) and x = -1 (for n = 6) leaves the cubic equation 8x^3 - 4x^2 - 4x + 1 = 0. This matches the required polynomial for the remaining three roots.

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