Expand ( (Squareroot of c ) -1) ^6

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baxthum8's profile pic

baxthum8 | High School Teacher | (Level 3) Associate Educator

Posted on

Expand `(sqrt(c)-1)^6`

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

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

From 2nd step replace to have:

`(c - 2csqrt(c) + 1) (c - 2csqrt(c) + 1) (c - 2csqrt(c) + 1)`

Use distributive property to get:

`c^3 - 6c^2sqrt(c) + 15c^2 - 20csqrt(c) + 14c - 6sqrt(c) + c + 1`

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

`(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`

Here we have used binomial theorem for expansion and

`^6C_r=(6!)/(r!(6-r)!)`

`sqrt(c)=c^(1/2)`

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