Expand `(sqrt(c-1))^6`

`(sqrt(c-1))^6 = (sqrt(c-1))^2*(sqrt(c-1))^2*(sqrt(c-1))^2`

Since `(sqrt(c-1))^2 = c-1`

We'd have: `(c-1)* (c - 1)* (c - 1)`

This is: `(c - 1) ^2 * (c - 1)`

`(c - 1)^2 = c^2 - 2c + 1`

Now, multiply `(c^2 - 2c + 1) (c - 1)`

`c^3 - c^2 - 2c^2 + 2c + 1c - 1`

**Finally we have:** `c^3 - 3c^2 + 3c - 1`

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