Solve {l}{M=sqrt[3]{8}*(-6)^2-(-5)^3{(3)}}{N=(-2)^3*(-2)^2-{(-3)}^3{(3)}^2-4}{a=M-N}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { M = \sqrt[3]{8} \cdot {(-6)} ^ {2} - {(-5)} ^ {3} {(3)} }\\ { N = {(-2)} ^ {3} \cdot {(-2)} ^ {2} - {(-3)} ^ {3} {(3)} ^ {2} - 4 }\\ { a = M - N }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for M, N, a, b

$$b=240$$

Solution Steps

Consider the first equation. Calculate \sqrt[3]{8} and get 2.

$$M=2\left(-6\right)^{2}-\left(-5\right)^{3}\times 3$$

Calculate -6 to the power of 2 and get 36.

$$M=2\times 36-\left(-5\right)^{3}\times 3$$

Multiply 2 and 36 to get 72.

$$M=72-\left(-5\right)^{3}\times 3$$

Calculate -5 to the power of 3 and get -125.

$$M=72-\left(-125\times 3\right)$$

Multiply -125 and 3 to get -375.

$$M=72-\left(-375\right)$$

The opposite of -375 is 375.

$$M=72+375$$

Add 72 and 375 to get 447.

$$M=447$$

Consider the second equation. To multiply powers of the same base, add their exponents. Add 3 and 2 to get 5.

$$N=\left(-2\right)^{5}-\left(-3\right)^{3}\times 3^{2}-4$$

Calculate -2 to the power of 5 and get -32.

$$N=-32-\left(-3\right)^{3}\times 3^{2}-4$$

Calculate -3 to the power of 3 and get -27.

$$N=-32-\left(-27\times 3^{2}\right)-4$$

Calculate 3 to the power of 2 and get 9.

$$N=-32-\left(-27\times 9\right)-4$$

Multiply -27 and 9 to get -243.

$$N=-32-\left(-243\right)-4$$

The opposite of -243 is 243.

$$N=-32+243-4$$

Add -32 and 243 to get 211.

$$N=211-4$$

Subtract 4 from 211 to get 207.

$$N=207$$

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

$$a=447-207$$

Subtract 207 from 447 to get 240.

$$a=240$$

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

$$b=240$$

The system is now solved.

$$M=447 N=207 a=240 b=240$$