Solve (3x-1/2)^3-27/8x^2(x-1)=1

$$( \frac { 3 x - 1 } { 2 } ) ^ { 3 } - \frac { 27 } { 8 } x ^ { 2 } ( x - 1 ) = 1$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

To raise \frac{3x-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x-1\right)^{3}}{2^{3}}-\frac{27}{8}x^{2}\left(x-1\right)=1$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(3x-1\right)^{3}.

$$\frac{27x^{3}-27x^{2}+9x-1}{2^{3}}-\frac{27}{8}x^{2}\left(x-1\right)=1$$

Calculate 2 to the power of 3 and get 8.

$$\frac{27x^{3}-27x^{2}+9x-1}{8}-\frac{27}{8}x^{2}\left(x-1\right)=1$$

Divide each term of 27x^{3}-27x^{2}+9x-1 by 8 to get \frac{27}{8}x^{3}-\frac{27}{8}x^{2}+\frac{9}{8}x-\frac{1}{8}.

$$\frac{27}{8}x^{3}-\frac{27}{8}x^{2}+\frac{9}{8}x-\frac{1}{8}-\frac{27}{8}x^{2}\left(x-1\right)=1$$

Use the distributive property to multiply -\frac{27}{8}x^{2} by x-1.

$$\frac{27}{8}x^{3}-\frac{27}{8}x^{2}+\frac{9}{8}x-\frac{1}{8}-\frac{27}{8}x^{3}+\frac{27}{8}x^{2}=1$$

Combine \frac{27}{8}x^{3} and -\frac{27}{8}x^{3} to get 0.

$$-\frac{27}{8}x^{2}+\frac{9}{8}x-\frac{1}{8}+\frac{27}{8}x^{2}=1$$

Combine -\frac{27}{8}x^{2} and \frac{27}{8}x^{2} to get 0.

$$\frac{9}{8}x-\frac{1}{8}=1$$

Add \frac{1}{8} to both sides.

$$\frac{9}{8}x=1+\frac{1}{8}$$

Add 1 and \frac{1}{8} to get \frac{9}{8}.

$$\frac{9}{8}x=\frac{9}{8}$$

Multiply both sides by \frac{8}{9}, the reciprocal of \frac{9}{8}.

$$x=\frac{9}{8}\times \frac{8}{9}$$

Multiply \frac{9}{8} and \frac{8}{9} to get 1.

$$x=1$$