Solve {l}{A={(2*{(-3)})}^3}{B={({(-3)}*5)}^2}{C={(4*{(-6)})}^3}{a=1}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { A = {(2 \cdot {(-3)})} ^ {3} }\\ { B = {({(-3)} \cdot 5)} ^ {2} }\\ { C = {(4 \cdot {(-6)})} ^ {3} }\\ { a = 1 }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for A, B, C, a, b

$$b=1$$

Solution Steps

Consider the first equation. Multiply 2 and -3 to get -6.

$$A=\left(-6\right)^{3}$$

Calculate -6 to the power of 3 and get -216.

$$A=-216$$

Consider the second equation. Multiply -3 and 5 to get -15.

$$B=\left(-15\right)^{2}$$

Calculate -15 to the power of 2 and get 225.

$$B=225$$

Consider the third equation. Multiply 4 and -6 to get -24.

$$C=\left(-24\right)^{3}$$

Calculate -24 to the power of 3 and get -13824.

$$C=-13824$$

The system is now solved.

$$A=-216 B=225 C=-13824 a=1 b=1$$