If $16a^4 + 36a^2b^2 + 81b^4 = 91$ and $4a^2 + 9b^2 - 6ab = 13$, then what is the value of $3ab$ ?

$-\frac{3}{2}$

If $16a^4 + 36a^2b^2 + 81b^4 = 91$

$4a^2 + 9b^2 - 6ab = 13$---(A)

Then what is the value of $3ab$ ?

x⁴ + x²y² + y⁴ = (x² – xy + y²) (x² + xy + y²)

$4a^2 + 9b^2 + 6ab = \frac{91}{13}$ = 7-----(B)

From A and B

12ab = -6

3ab = \(\frac{-6 \times 3}{12}\)

3ab = $-\frac{3}{2}$