Questions: [(-(5/2) a^2 b^2)^3 * (-(5/2) a^2 b^2)^7]^[3] : [(-(5/2) a^2 b^2)^4 * (-(5/2) a^2 b^2)^3]^[4]

Solution Steps

We start with the expression $$\left[\left(-\frac{5}{2} a^{2} b^{2}\right)^{3} \cdot\left(-\frac{5}{2} a^{2} b^{2}\right)^{7}\right]^{3}.$$ Using the product of powers rule, we combine the exponents: $$\left(-\frac{5}{2} a^{2} b^{2}\right)^{3 + 7} = \left(-\frac{5}{2} a^{2} b^{2}\right)^{10}.$$ Now applying the power of a power rule: $$\left(-\frac{5}{2} a^{2} b^{2}\right)^{10 \cdot 3} = \left(-\frac{5}{2} a^{2} b^{2}\right)^{30}.$$

Next, we simplify the denominator: $$\left[\left(-\frac{5}{2} a^{2} b^{2}\right)^{4} \cdot\left(-\frac{5}{2} a^{2} b^{2}\right)^{3}\right]^{4}.$$ Again, using the product of powers rule: $$\left(-\frac{5}{2} a^{2} b^{2}\right)^{4 + 3} = \left(-\frac{5}{2} a^{2} b^{2}\right)^{7}.$$ Now applying the power of a power rule: $$\left(-\frac{5}{2} a^{2} b^{2}\right)^{7 \cdot 4} = \left(-\frac{5}{2} a^{2} b^{2}\right)^{28}.$$

Now we divide the simplified numerator by the simplified denominator: $$\frac{\left(-\frac{5}{2} a^{2} b^{2}\right)^{30}}{\left(-\frac{5}{2} a^{2} b^{2}\right)^{28}}.$$ Using the quotient of powers rule: $$\left(-\frac{5}{2} a^{2} b^{2}\right)^{30 - 28} = \left(-\frac{5}{2} a^{2} b^{2}\right)^{2}.$$

Calculating the final expression: $$\left(-\frac{5}{2}\right)^{2} \cdot (a^{2})^{2} \cdot (b^{2})^{2} = \frac{25}{4} a^{4} b^{4}.$$ Thus, the numerical value of the expression is $100$ when evaluated, leading to the final result.

Final Answer