# Expand (√c-1)^6

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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`

`(sqrt(c)-1)^6=sum_(r=0)^6(^6C_r(sqrt(c))^(6-r)(-1)^r)`

`=^6C_0(sqrt(c))^6(-1)^0+^6C_1(sqrt(c))^5(-1)^1+^6C_2(sqrt(c))^4(-1)^2+`

`^6C_3(sqrt(c))^3(-1)^3+^6C_4(sqrt(c))^2(-1)^4+^6C_5(sqrt(c))^1(-1)^5+`

`^6C_6(sqrt(c))^0(-1)^6`

`=c^(6/2)-6c^(5/2)+15c^(4/2)-20c^(3/2)+15c^(2/2)-6c^(1/2)+1`

`=c^3-6c^2sqrt(c)+15c^2-20csqrt(c)+15c-6sqrt(c)+1`

`` We have used binomial expansion for solving this problem

and `^6C_r=(6!)/(r!(6-r)!)` ,also `sqrt(c)=c^(1/2)` etc.