# Please help me solve (-216^6) ^1/3   2/7 ```root3(-343)` - 1/4 ```root3(32)` + 1/2 `root3(108)`

Asked on by monique06

crmhaske | College Teacher | (Level 3) Associate Educator

Posted on

The solution to the first problem will depend on the location of the brackets.  If it is correct as written:

Assuming you meant to include the negative sign in brackets to the power of six:

`((-216)^6)^(1/3)=45656`

Next, we'll compute the three terms separately first:

`2/7(-343)^(1/3)=(2/7)(-7)=-2`

`1/4(32)^(1/3)=(1/4)(8)^(1/3)(4)^(1/3)=(1/2)4^(1/3)`

`(1/2)(108)^(1/3)=(1/2)(27)^(1/3)(4)^(1/3)=(3/2)4^(1/3)`

`-2-(1/2)4^(1/3)+(3/2)4^(1/3)`

`=-2+4^(1/3)`

Sources:

oldnick | (Level 1) Valedictorian

Posted on

So:

`2/7 root(3)(343)-1/4 root(3)(32)+1/2 root(3)(108)=` `-2-1/2root(3)(4)+3/2 root(3)(4)=`

`` `=root(3)(4)-2`

oldnick | (Level 1) Valedictorian

Posted on

`A)`    `((-216)^6)^(1/3)= (-216)^(6 xx 1/3)=(-216)^2=((-6)^3)^2=`

`=(-6)^6`

`B)`     `2/7 xx root(3)(-343)= 2/7 xx root(3)(-7^3)=2/7 xx (-7)^(3xx1/3)=` `2/7 xx (-7)=-2`

`C)`    `(-1/4)root(3)(32)=-1/4 (2^5)^(1/3)=` `(-1/4) 2^(5/3)=-1/4 2^((3+2)/3)=`

`=(-1/4) 2^(3/3 + 2/3)=` `(-1/4) xx 2 xx2^(2/3)=` `-1/2 root(3)(4)`

`D)`    `1/2 root(3)(108)=` `1/2 root(3)(27 xx 4)=1/2 root(3)(3^3 xx 4)=3/2 root(3)(4)`

TO BE CONTINUE...............

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