Question : If for a non-zero $x$, $3x^{2}+5x+3=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Option 1: $\frac{10}{27}$
Option 2: $-\frac{10}{27}$
Option 3: $\frac{2}{3}$
Option 4: $-\frac{2}{3}$
Correct Answer: $\frac{10}{27}$
Solution :
Given: $3x^{2}+5x+3=0$
$3x^2 + 3=-5x$
Dividing by $3x$ on both sides,
$x+\frac{1}{x} = -\frac{5}{3}$
As $(a+b)^3 = a^3 + b^3 +3ab(a+b)$
$\therefore$ $x^3 + \frac{1}{x^3} = (\frac{-5}{3})^3 - 3\times \frac{-5}{3}$
= $\frac{-125}{27}+5$
= $\frac{–125+135}{27}=\frac{10}{27}$
Hence, the correct answer is $\frac{10}{27}$.
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