Question #d8979

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Answer 1 · 2017-10-15T11:32:37

#322#

I shall assume you mean:

If #x+1/x=3#, then what is #x^6+1/x^6#?

Notice the following:

#(x+1/x)^2=3^2#

#x^2+2*x*1/x+(1/x)^2=9#

#x^2+2+1/x^2=9#

#x^2+1/x^2=7#

We can then cube both sides to get the power of #6# we are looking for.

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

#(x^2)^3+3*(x^2)^2*1/x^2+3*x^2*(1/x^2)^2+(1/x^2)^3=343#

#x^6+3x^2+3*1/x^2+1/x^6=343#

#x^6+1/x^6+3(x^2+1/x^2)=343#

However, if you remember from before, we know that #x^2+1/x^2=7#. We can plug that in to this equation:

#x^6+1/x^6+3(7)=343#

#x^6+1/x^6=343-21#

#x^6+1/x^6=322#