If #f(x)# = #x^3# - #3x# + #3x# then #f#(#root(3)#7 + #1#) equals what?

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Answer 1 · 2018-06-18T13:31:44

#8+3(7^(2/3)+7^(1/3))#

Given: #f(x)=x^3-3x+3x#.

We can simplify the function into just:

#f(x)=x^3#

When #x=root(3)7+1#, the function equals to:

#f(x)=(root(3)7+1)^3#

Now, we use the fact that #(a+b)^3=a^3+3a^2b+3ab^2+b^3#. Letting #a=root(3)7# and #b=1#, we get:

#=(root(3)7)^3+3(root(3)7)^2*1+3root(3)7*1^2+1^3#

#=7+3*7^(2/3)+3*7^(1/3)+1#

#=8+3(7^(2/3)+7^(1/3))#