Solve {l}{A=-(-2)^3+3{(5^2-10*2)}+(-1)^20100}{a=2-(-3)^2}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { A = -{(-2)} ^ {3} + 3 {(5 ^ {2} - 10 \cdot 2)} + {(-1)} ^ {2010} 0 }\\ { a = 2 - {(-3)} ^ {2} }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for A, a, b

$$b=-7$$

Solution Steps

Consider the first equation. Calculate -2 to the power of 3 and get -8.

$$A=-\left(-8\right)+3\left(5^{2}-10\times 2\right)+\left(-1\right)^{2010}\times 0$$

The opposite of -8 is 8.

$$A=8+3\left(5^{2}-10\times 2\right)+\left(-1\right)^{2010}\times 0$$

Calculate 5 to the power of 2 and get 25.

$$A=8+3\left(25-10\times 2\right)+\left(-1\right)^{2010}\times 0$$

Multiply 10 and 2 to get 20.

$$A=8+3\left(25-20\right)+\left(-1\right)^{2010}\times 0$$

Subtract 20 from 25 to get 5.

$$A=8+3\times 5+\left(-1\right)^{2010}\times 0$$

Multiply 3 and 5 to get 15.

$$A=8+15+\left(-1\right)^{2010}\times 0$$

Add 8 and 15 to get 23.

$$A=23+\left(-1\right)^{2010}\times 0$$

Calculate -1 to the power of 2010 and get 1.

$$A=23+1\times 0$$

Multiply 1 and 0 to get 0.

$$A=23+0$$

Add 23 and 0 to get 23.

$$A=23$$

Consider the second equation. Calculate -3 to the power of 2 and get 9.

$$a=2-9$$

Subtract 9 from 2 to get -7.

$$a=-7$$

Consider the third equation. Insert the known values of variables into the equation.

$$b=-7$$

The system is now solved.

$$A=23 a=-7 b=-7$$