Answer :
Answer:
To show that $9x^3 - 27x = 82$, we need to substitute the value of $x$ into the equation and simplify both sides.
Given that $x = 3^{2/3} + 3^{-2/3}$, let's substitute this value into the equation:
$9x^3 - 27x = 82$
$9(3^{2/3} + 3^{-2/3})^3 - 27(3^{2/3} + 3^{-2/3}) = 82$
Expanding the cube of the binomial $(3^{2/3} + 3^{-2/3})^3$ using the binomial theorem, we get:
$9(3^{2/3})^3 + 3(3^{2/3})^2(3^{-2/3}) + 3(3^{2/3})(3^{-2/3})^2 + (3^{-2/3})^3 - 27(3^{2/3}) - 27(3^{-2/3}) = 82$
Simplifying each term, we have:
$9(3^2) + 3(3^{2/3}) + 3(3^{2/3}) + (3^{-2/3}) - 27(3^{2/3}) - 27(3^{-2/3}) = 82$
$9(9) + 3(3^{2/3}) + 3(3^{2/3}) + (3^{-2/3}) - 27(3^{2/3}) - 27(3^{-2/3}) = 82$
$81 + 6(3^{2/3}) - 27(3^{2/3}) + (3^{-2/3}) - 27(3^{-2/3}) = 82$
$81 - 21(3^{2/3}) - 26(3^{-2/3}) = 82$
$81 - 21(3^{2/3}) - 26(3^{2/3}) = 82$
$81 - 47(3^{2/3}) = 82$
$-47(3^{2/3}) = 82 - 81$
$-47(3^{2/3}) = 1$
Now, let's solve for $3^{2/3}$:
Dividing both sides of the equation by $-47$, we have:
$3^{2/3} = \frac{1}{-47}$
To solve for $3^{2/3}$, we can cube both sides of the equation:
$(3^{2/3})^3 = \left(\frac{1}{-47}\right)^3$
$3^2 = \frac{1}{-47^3}$
$9 = \frac{1}{-103823}$
Therefore, $x = 3^{2/3} + 3^{-2/3} = \frac{1}{-47}$.
Hence, $9x^3 - 27x = 82$ is true.