Solve {l}{a=1/3}{b={(3a+1)}{(9a^2-3a+1)}}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { a = \frac{1}{3} }\\ { b = {(3 a + 1)} {(9 a ^ {2} - 3 a + 1)} }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for a, b, c

$$c=2$$

Solution Steps

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

$$b=\left(3\times \frac{1}{3}+1\right)\left(9\times \left(\frac{1}{3}\right)^{2}-3\times \frac{1}{3}+1\right)$$

Multiply 3 and \frac{1}{3} to get 1.

$$b=\left(1+1\right)\left(9\times \left(\frac{1}{3}\right)^{2}-3\times \frac{1}{3}+1\right)$$

Add 1 and 1 to get 2.

$$b=2\left(9\times \left(\frac{1}{3}\right)^{2}-3\times \frac{1}{3}+1\right)$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$b=2\left(9\times \frac{1}{9}-3\times \frac{1}{3}+1\right)$$

Multiply 9 and \frac{1}{9} to get 1.

$$b=2\left(1-3\times \frac{1}{3}+1\right)$$

Multiply -3 and \frac{1}{3} to get -1.

$$b=2\left(1-1+1\right)$$

Subtract 1 from 1 to get 0.

$$b=2\times 1$$

Multiply 2 and 1 to get 2.

$$b=2$$

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

$$c=2$$

The system is now solved.

$$a=\frac{1}{3} b=2 c=2$$