Problem based on Algebraic identities Algebraic identities:
$$(a+b)^2 = a^2 + b^2 + 2ab$$
$$(a-b)^2 = a^2 + b^2 - 2ab$$
Other identities can also come to solve this question?
If $x + 1/x = 5$ and $x^2 + 1/x^3 = 8$, then what would be the value of $x^3 + 1/x^2$?
Possible answers:
a- 215
b- 125
c- 256
d- 525
 A: let $$a=x^2+\dfrac{1}{x^3},b=x^3+\dfrac{1}{x^2}$$
then $$a+b=x^2+\dfrac{1}{x^2}+x^3+\dfrac{1}{x^3}$$
and
$$x^2+\dfrac{1}{x^2}=(x+\dfrac{1}{x})^2-2=23$$
$$x^3+\dfrac{1}{x^3}=(x+\dfrac{1}{x})^3-3(x+\dfrac{1}{x})=125-15=110$$
so
$$a+b=133$$
then
$$b=133-8=125$$
A: The premise of the question is wrong.  Neither of the two solutions of $x + 1/x = 5$ satisfy $x^2 + 1/x^3 = 8$.  Nor do they make $x^3 + 1/x^2$ equal to any of (a), (b), (c), (d).  
Since "A implies B" is true whenever A is false, technically all four answers are correct.
A: One more useful identity is $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. The answer is 125.
A: we have $x^2+1=5x$ and $x^5+1=8x^3$
$$\frac{x^5+1}{x^2}=\frac{8x^3}{x^2}=8x$$
We need to solve for $x$  from $x^2-5x+1=0$ which will satisfy  $x^5+1=8x^3$ as well
A: see ($x^2$+$x^-3$)($x^3$+$x^-2$)


*

*=$x^5$+$x^-5$+2

*NOW x+$\frac{1}{x}$=5. FROM THIS WE GET $x^2$+$x^-2$=23 and $x^3$+$x^-3$=110

*NOW $x^5$+$x^-5$=($x^2$+$x^-2$)($x^3$+$x^-3$)$-$(x+$x^-1$)=23.110$-$5=2525

*SO($x^2$+$x^-3$)($x^3+$$x^-2$)=2525+2=2527

*so $x^3$+$x^-2$=$\frac{2527}{8}${as $x^2$+$x^-3$=8}

