If a constant F of (3a3b2c⁻²)2 Newtons moves an object of a distance D equivalent to (2a2b⁻³c2)3 metres, find the work done in metric?

Work done ($W$) is calculated as the product of force ($F$) and distance ($D$).

$$W = F \times D$$

The given force is $F = (3a^3b^2c^{-2})^2$ Newtons.
To simplify, apply the power rule $(xy)^n = x^n y^n$ and $(x^m)^n = x^{mn}$:

$$F = 3^2 \cdot (a^3)^2 \cdot (b^2)^2 \cdot (c^{-2})^2$$ $$F = 9 \cdot a^{3 \times 2} \cdot b^{2 \times 2} \cdot c^{-2 \times 2}$$ $$F = 9a^6b^4c^{-4} \text{ Newtons}$$

The given distance is $D = (2a^2b^{-3}c^2)^3$ metres.
To simplify, apply the power rule $(xy)^n = x^n y^n$ and $(x^m)^n = x^{mn}$:

$$D = 2^3 \cdot (a^2)^3 \cdot (b^{-3})^3 \cdot (c^2)^3$$ $$D = 8 \cdot a^{2 \times 3} \cdot b^{-3 \times 3} \cdot c^{2 \times 3}$$ $$D = 8a^6b^{-9}c^6 \text{ metres}$$

Now, multiply the simplified force and distance expressions:

$$W = (9a^6b^4c^{-4}) \times (8a^6b^{-9}c^6)$$

Multiply the numerical coefficients:

$$W = (9 \times 8) \cdot (a^6 \cdot a^6) \cdot (b^4 \cdot b^{-9}) \cdot (c^{-4} \cdot c^6)$$

Apply the exponent rule $x^m \cdot x^n = x^{m+n}$:

$$W = 72 \cdot a^{6+6} \cdot b^{4+(-9)} \cdot c^{-4+6}$$ $$W = 72 \cdot a^{12} \cdot b^{4-9} \cdot c^{2}$$ $$W = 72a^{12}b^{-5}c^2 \text{ Newton-metres}$$

The work done is $\text{}\mathbf{72a^{12}b^{-5}c^2}\text{ Newton-metres}$