Solve y=1/8(x-1)^3(x+3)^3

$$y = \frac { 1 } { 8 } ( x - 1 ) ^ { 3 } ( x + 3 ) ^ { 3 }$$

Solve for y

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

Solution Steps

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

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

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

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

Use the distributive property to multiply \frac{1}{8} by x^{3}-3x^{2}+3x-1.

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

Use the distributive property to multiply \frac{1}{8}x^{3}-\frac{3}{8}x^{2}+\frac{3}{8}x-\frac{1}{8} by x^{3}+9x^{2}+27x+27 and combine like terms.

$$y=\frac{1}{8}x^{6}+\frac{3}{4}x^{5}-\frac{7}{2}x^{3}+\frac{3}{8}x^{4}+\frac{27}{4}x-\frac{9}{8}x^{2}-\frac{27}{8}$$