Solve C=(2x+1)^3-8

$$C = ( 2 x + 1 ) ^ { 3 } - 8$$

Solve for C

$$C=\left(2x-1\right)\left(4x^{2}+8x+7\right)$$

Solution Steps

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

$$C=8x^{3}+12x^{2}+6x+1-8$$

Subtract 8 from 1 to get -7.

$$C=8x^{3}+12x^{2}+6x-7$$

Solve for x (complex solution)

$$x=\frac{\sqrt[3]{C+8}-1}{2}$$

$$x=\frac{e^{\frac{4\pi i}{3}}\sqrt[3]{C+8}-1}{2}$$

$$x=\frac{e^{\frac{2\pi i}{3}}\sqrt[3]{C+8}-1}{2}$$

Solve for x

$$x=\frac{\sqrt[3]{C+8}-1}{2}$$