1. **Problem:** Calculate the value of $$\left[ \left(4^{-1} + 3^{-1}\right) + \left(\frac{7}{9}\right)^{-1} \right]^{-3}$$. 2. **Recall the rules:** - Negative exponent rule: $$a^{-n} = \frac{1}{a^n}$$. - Inverse of a fraction: $$\left(\frac{a}{b}\right)^{-1} = \frac{b}{a}$$. - Powers of sums require careful addition inside parentheses before applying exponents. 3. **Calculate each term inside the brackets:** - $$4^{-1} = \frac{1}{4}$$ - $$3^{-1} = \frac{1}{3}$$ - $$\left(\frac{7}{9}\right)^{-1} = \frac{9}{7}$$ 4. **Sum the terms inside the brackets:** $$\frac{1}{4} + \frac{1}{3} + \frac{9}{7}$$ 5. **Find common denominator for $$\frac{1}{4}$$ and $$\frac{1}{3}$$:** $$\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}$$ 6. **Add $$\frac{7}{12}$$ and $$\frac{9}{7}$$:** Common denominator is $$12 \times 7 = 84$$ $$\frac{7}{12} = \frac{7 \times 7}{84} = \frac{49}{84}$$ $$\frac{9}{7} = \frac{9 \times 12}{84} = \frac{108}{84}$$ Sum: $$\frac{49}{84} + \frac{108}{84} = \frac{157}{84}$$ 7. **Apply the outer exponent $$-3$$:** $$\left(\frac{157}{84}\right)^{-3} = \left(\frac{84}{157}\right)^3 = \frac{84^3}{157^3}$$ 8. **Final answer:** $$\boxed{\frac{84^3}{157^3}}$$ This is the exact value of the expression.