Solve ((7×3)²)⁶ + (3⁻¹)³×(2³)⁴: Complex Exponent Challenge

Solve the following problem:

$((7\times3)^2)^6+(3^{-1})^3\times(2^3)^4=$

Let's handle each expression in the problem separately:

a. We'll start with the leftmost expression, first calculating the result of the multiplication in parentheses, and then using the power rule for power to a power:

$(a^m)^n=a^{m\cdot n}$

Let's apply this to the problem for the first expression from the left:

$((7\cdot3)^2)^6=(21^2)^6=21^{2\cdot6}=21^{12}$

In the final step we calculated the result of multiplication in the power expression.

We're now finished with this expression, let's move on to the next expression from the left.

b. Continue with the second expression from the left, using the power rule for power to a power that we mentioned above and apply it separately to each factor in this expression:

$(3^{-1})^3\cdot(2^3)^4=3^{-1\cdot3}\cdot2^{3\cdot4}=3^{-3}\cdot2^{12}$

Note that the multiplication factors that we obtained have different bases, thus we cannot further simplify this expression,

Therefore, let's combine parts a and b above in the result of the original problem:

$((7\cdot3)^2)^6+(3^{-1})^3\cdot(2^3)^4=21^{12}+3^{-3}\cdot2^{12}$

The correct answer is answer d.