Question 353546
PLEASE HELP ME TO SOLVE THESE QUESTIONS.............

Q1.IF {{{x^3+1/x^3=2}}} then find the value of x+1/x
<pre>
Using the binomial theorem:

{{{(x+1/x)^3=x^3+3x^2(1/x)+3x(1/x)^2+(1/x)^3)}}}

{{{(x+1/x)^3=x^3+3x+3x(1/x^2)+1/x^3}}}

{{{(x+1/x)^3=x^3+3x+3/x+1/x^3}}}

{{{(x+1/x)^3=x^3+1/x^3+3x+3/x}}}

{{{(x+1/x)^3=(x^3+1/x^3)+3(x+1/x)}}}

and since we are given {{{x^3+1/x^3=2}}},

{{{(x+1/x)^3=2+3(x+1/x)}}}

{{{(x+1/x)^3-3(x+1/x)-2=0}}}

Let {{{y = x+1/x}}}, which is what we want to find.

{{{y^3-3y-2=0}}}

Feasible rational roots are ±1, ±2
Using synthetic division we find that -1 is a solution

-1|1  0 -3 -2
  |<u>  -1  1  2</u>
   1 -1 -2  0

Thus we have factored

{{{y^3-3y-2=0}}}

as

{{{(y+1)(y^2-y-2)=0}}}

Factoring further:

{{{(y+1)(y+1)(y-2)=0}}}

Thus the solutions are y = -1, y = 2

so {{{x+1/x}}} is either -1 or 2.

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</pre>
Q2.FIND THE VALUE OF {{{(x+y)^3+(x-2y+2)}}}

{{{(x+y)^3+(x-2y+2)}}}
<pre>
Something is missing from this one that must be given 
to find a solution.  Check the problem again and repost 
it with the part that's missing here.

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</pre>
Q3.IF {{{(x+1/x)*(x-1/x)*(x^2+1/x^2)*(x^4+1/x^4)*(x^8+1/x^8)=3/2}}} .then find the value of {{{x^16}}}
<pre>
{{{(x+1/x)*(x-1/x)*(x^2+1/x^2)*(x^4+1/x^4)*(x^8+1/x^8)=3/2}}}

Multiplying the first two factors:

{{{(x^2-1/x^2)*(x^2+1/x^2)*(x^4+1/x^4)*(x^8+1/x^8)=3/2}}}

Multiplying the first two factors:

{{{(x^4-1/x^4)*(x^4+1/x^4)*(x^8+1/x^8)=3/2}}}

Multiplying the first two factors:

{{{(x^8-1/x^8)*(x^8+1/x^8)=3/2}}}

Multiplying those factors:

{{{x^16-1/x^16=3/2}}}

Let {{{y = x^16}}}

{{{y-1/y=3/2}}}

Clear of fractions:

{{{2y^2-2=3y}}}

{{{2y^2-3y-2=0}}}

Factoring,

{{{(y-2)(2y+1)=0}}}

Solutions are {{{y = 2}}}, {{{y=-1/2}}}

and since {{{y=x^16}}}

{{{x^16=2}}} or {{{x^16=-1/2}}}

(In the case of the second negative solution, 
x would have to be imaginary).

Edwin</pre>