How do you multiply #x^(2/3)(x^(1/4) - x) #?

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Answer 1 · 2017-02-01T08:31:16

#x^(11/12)-x^(5/3)#

We have:

#x^(2/3)(x^(1/4)-x)#

I'm first going to rewrite this so that we can see the exponent on the "plain" #x# term: #x=x^1 =>#

#x^(2/3)(x^(1/4)-x^1)#

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Before we move on, it's important to see that #(x^(1/4)-x^1)!=x^(-3/4)#

For instance, if we set #x=16 => (16^(1/4)-16^1)!=16^(-3/4)=>(2-16)!=-1/8#

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We can use the rule #x^a xx x^b=x^(a+b)#:

#x^(2/3)(x^(1/4)-x^1)#

#x^(1/4+2/3)-x^(1+2/3)#

#x^(3/12+8/12)-x^(3/3+2/3)#

#x^(11/12)-x^(5/3)#